CN204284427U - Heating self-reinforcing pressure vessel in compression elasticity temperature difference prestressing force in a kind of - Google Patents

Abstract

In a kind of, heating self-reinforcing pressure vessel in compression elasticity temperature difference prestressing force, solves and improves the technical problem such as pressurized container bearing capacity and Security.The temperature difference stress counteracting part that this kind of pressurized container utilizes inside and outside wall temperature difference to produce is by the mechanical stress operating interior pressure p and cause, to reduce and its stress distribution of homogenizing, thus strengthen its bearing capacity,, saving more safer, convenient, reliable than mechanical prestress method self-reinforcing pressure vessel, designs more flexible.Its drip irrigation device is: with critical temperature difference or the desirable temperature difference control inside and outside wall temperature difference, with or about beam diameter is than the size between k, p, E, a, u, temperature difference dt; Wherein E be the Young's modulus of container material, a thermal expansion coefficient that is container material, u be container material Poisson's ratio, s _yfor yield strength, the k=r of container material _o/ r _i, r _ifor container inner wall radius surface, r _ofor container outer wall radius surface.The footpath of this kind of pressurized container reaches than minimum or , most high bearing capacity reaches p/s _y=ln k or p/s _y=(k ²-1)/k ².

Description

Heating self-reinforcing pressure vessel in compression elasticity temperature difference prestressing force in a kind of

Technical field

The utility model relates to heating self-reinforcing pressure vessel in a kind of interior compression elasticity temperature difference prestressing force, relates to the fields such as technology such as machinery, chemical industry, pharmacy, the energy, material, food, metallurgy, oil, building, Aeronautics and Astronautics, weapons.

Background technique

Pressurized container is key, the special equipment of many industrial departments, is widely used in all trades and professions, as departments such as machinery, chemical industry, pharmacy, the energy, material, food, metallurgy, oil, building, Aeronautics and Astronautics, weapons.The main body overwhelming majority of pressurized container is cylinder, and when cylinder bears working pressure, the stress in its wall is very uneven, as shown in Figure of description 5.Container thickness is larger, and stress is more uneven.If press maximum stress design pressure container, its wall thickness can be made very large, and wall thickness greatly not only wastes material, resource, fund, increases cost, also has potential safety hazard.Must manage to reduce the stress in container wall, the intensity of container could be improved, save material, reduce costs, and improve the Security of pressurized container.The method reducing the stress in container wall has a lot, self intensification technology be namely reduce and average stress, raising pressurized container bearing capacity and Security thereof important and effective means.The concrete grammar of pressurized container self intensification has again a lot, such as, before container comes into operation, larger mechanical pressure is applied to it, its internal layer portion of material is made to produce plastic deformation, outer section material is still elastic state, after removal mechanical pressure, just in container wall, produce mechanical prestress (residual stress): internal layer portion of material is pressure stress, outer section material is tensile stress.The stress that mechanical prestress and container operation pressure cause is superimposed can reduce stress.The utility model then adopts temperature difference prestressing force self-strengthening method to construct heating self-reinforcing pressure vessel in a kind of interior compression elasticity temperature difference prestressing force.The stress that namely described temperature difference prestressing force caused by the inside and outside wall surface temperature difference of container due to the reason of metal heat-expansion shrinkage, temperature difference is caused by the operational difference of container itself, or heats its inside and outside wall or cool and cause before container comes into operation.Temperature difference prestressed pressure container is more safer than mechanical prestress pressurized container, convenient, reliable, save, flexibly, because (1) produce prestressed method by temperature difference stress to there is not medium of exerting pressure, therefore not dangerous, do not need expensive hydraulic press etc. to exert pressure equipment yet; (2) control of the temperature difference is comparatively easy, and thus the size of temperature difference stress and uniformity thereof are easily guaranteed; (3) once temperature difference stress is excessive, owing to there is not pressure medium, therefore unlikelyly as mechanical stress, the catastrophic failures such as pressurized container blast are caused; (4) designer studies discovery, the prestressed size of the temperature difference and the regularity of distribution and the temperature difference dtbe closely connected, therefore, can change according to operational condition dtto obtain different operational stresses induced states, visual cell structure optimization space is very large, and this just can obtain the design proposal of maneuverability.

Summary of the invention

The purpose of this utility model is to provide heating self-reinforcing pressure vessel in a kind of interior compression elasticity temperature difference prestressing force.

The technological scheme that the utility model adopts is: heating self-reinforcing pressure vessel in compression elasticity temperature difference prestressing force in structure is a kind of, it is characterized in that: the inside and outside wall of described pressurized container exists the temperature difference, the temperature difference makes pressurized container wall produce temperature difference prestressing force, total stress during described pressurized container work is that temperature difference prestressing force and mechanical stress are superimposed, described mechanical stress finger pressure container work pressure pthe stress produced, described pressurized container working pressure pfor interior pressure; Described internal pressure vessel walls radius surface is r _i, outer wall radius is r _o, internal face temperature is t _i, outer wall temperature is t _o, the temperature difference dt= t _i- t _o, t _i> t _o, be called interior heating; Described pressurized container, under the effect of temperature difference prestressing force, is in elastic state, namely dt≤ dt _c, dt _cfor temperature difference prestressing force has just made internal pressure vessel walls face produce the temperature difference of surrender, be called critical temperature difference, , wherein efor container material Young's modulus, afor container material thermal expansion coefficient, ufor container material Poisson's ratio, s _yfor container material yield strength, kfor the footpath ratio of container, k= r _o/ r _i.The described temperature difference is caused by the operating temperature of pressurized container itself, or heats the inside and outside wall of pressurized container or cool and cause. k>= k _ctime, its temperature difference is defined as dt≤ dt _c, bearing capacity p/ s _ywith regulation; dt= dt _ctime, bearing capacity is , now container footpath ratio is ; Wherein p _efor initial yield mechanical load or maximum machine elastic load, k _cfor by formula determine kvalue, is called critical footpath ratio, k _c=2.2184574899167 ..., all the other symbols are the same. k≤ k _ctime, its temperature difference is defined as dt≤ dt _x, bearing capacity p/ s _ywith regulation; dt= dt _xtime, bearing capacity is p= s _yln k= p _y, now container footpath ratio is ; Wherein , be called the desirable temperature difference, p _yfor entirely surrendering mechanical load or integral yield mechanical load, all the other symbols are the same. p≤ p _cor p/ s _y≤ p _c/ s _ytime, its footpath ratio is ; p>= p _cor p/ s _y>= p _c/ s _ytime, its footpath ratio is ; Wherein p _cfor by formula determine pvalue, p _c/ s _y=0.796812 ..., all the other symbols are the same.The physical dimension of this pressurized container, i.e. footpath ratio k, bearing capacity p/ s _y, material property factor ( e, a, u, s _y), the temperature difference dtbetween with or relation constraint, container wall thickness b= r _o- r _i= r _i( k-1); k, p, e, a, u, s _y, dtsize adjust according to actual needs, combine, mate; Each symbolic significance is the same.

The beneficial effects of the utility model and advantage are: poor in pressurized container inside and outside wall formation temperature, the temperature difference stress produced by the temperature difference offsets the mechanical stress of a part caused by its operation pressure, with its stress distribution of homogenizing, thus strengthen its bearing capacity, to form a kind of temperature difference prestressing force self-reinforcing pressure vessel.For making constructed pressurized container science, advanced person, to reach optimum efficiency, the utility model provides the footpath ratio of constructed pressurized container k, bearing capacity p/ s _y, material property factor ( e, a, u, s _y), the temperature difference dtbetween the concrete technological scheme such as the constraint rule that should follow, as critical temperature difference, the desirable temperature difference, best footpath ratio with deng.All the other beneficial effects and advantage are as described in " background technique " end.

Accompanying drawing explanation

Fig. 1 is the cylindrical pressure vessel plan view that inside and outside wall exists the temperature difference.

Fig. 2 is the cylindrical pressure vessel plan view that inside and outside wall exists the temperature difference.

Elasticity temperature difference stress distribution map when Fig. 3 is typical interior heating.

Fig. 4 is k~ i, k~ n, k~ z, k~ m, k~ wgraph of a relation.

Fig. 5 is k=2, dt= dt _x=124.411 ° of C, p/ s _y=ln ktotal stress distribution map when=0.693147391.

Fig. 6 is k>= k ₁>=1,1≤ lwhen≤2 m, wcurve.

Fig. 7 is k ₁=1.671272, lwhen=1.6 m, wcurve.

Embodiment

Provide the scientific basis constructing this self-reinforcing pressure vessel below, and provide enforcement example in conjunction with these theories.

The main body overwhelming majority of pressurized container is cylinder, if the inside and outside radius of cylindrical pressure vessel is respectively r _i, r _o, the symbol of subscripting i, o represents the value on inside and outside wall respectively; Inside and outside wall temperature is respectively t _i, t _o.The arbitrary radius of barrel is rthe temperature at place is t, see Figure of description 1.By analysis, studying, there is the temperature difference in the inside and outside wall of cylindrical pressure vessel dttime, in wall, (radius is at any point place r) temperature difference stress be:

(1)

Wherein, s _r ^t, s _t ^t, s _z ^t-radial direction, hoop, axial temperature difference stress, MPa;

p _t-thermal force, Mpa, ;

ethe Young's modulus of elasticity of-pressure vessel material, Mpa;

athe thermal expansion coefficient of-pressure vessel material, ° C ^-1;

uthe Poisson's ratio of-pressure vessel material, dimensionless;

dtthe temperature difference of the inside and outside wall of-pressurized container, ° C, dt= t _i- t _o;

kthe footpath ratio of-pressurized container, dimensionless, k= r _o/ r _i;

k _r- k _r= r _o/ r, dimensionless;

x-relative position, x= r/ r _i, dimensionless.

No matter interior heating or external heat, internal face stress absolute value is greater than outer wall stress absolute value, and s _ri ^t=0, therefore, along with | dt| increase, always internal face is first surrendered.

[example]: the Young's modulus of general steel is e=1.95 × 10 ⁵mPa, Poisson's ratio is u=0.3, thermal expansion coefficient is a=1.2 × 10 ^-5° C ^-1if, dtheating in=50 ° of C(), then p _t=83.57143MPa.If r _i=200mm, r _o=500mm, namely k=2.5, its temperature difference stress is as shown in Figure of description 3.

Fig. 3 be typical in add thermoelasticity temperature difference stress, no matter the temperature difference is much, de-stress numerical value have different outside, temperature difference stress distribution pattern is the same with Fig. 3: closely internal face place, s _t ^t, s _z ^tfor pressure stress, nearly outer wall place, s _t ^t, s _z ^tfor tensile stress.This distribution just in time presses the mechanical stress caused in homogenizing.During interior press operation, interior heating is favourable to reduction stress; External heat will worsen stress.Therefore, the utility model specially discusses the technical measures of interior pressure, interior heating.

When s _ri ^t- s _zi ^t= s _ytime (interior heating), container inner wall face start surrender, by formula (1) the internal face initial yield temperature difference, be called critical temperature difference dt _c:

(interior heating) (2)

Wherein s _yfor the yield strength of pressure vessel material, . dt< dt _ctime, temperature difference stress does not cause container to produce surrender.Container wall is thicker, and it is less that internal face starts to surrender the required temperature difference. k=1， i=1； k→∞， i→0.5。Thermal force when surrendering by formula (2) obtains internal face, is called critical heat load (interior heating)

Temperature difference stress is called total stress with superposing of mechanical stress, and when cylindrical pressure vessel is by interior pressure, total stress is

(3)

In formula, s _r, s _t, s _z-radial direction, hoop, axial total stress, MPa.

(4)

Internal face, x=1: (5)

Outer wall, x= k: (6)

Order s _ti- s _ri≤ s _yobtain bearing capacity p/ s _y:

(7)

Or (7a)

Formula (7) shows, dtlarger, bearing capacity or allowable load p ₁/ s _ylarger.

Order s _to- s _ro≤ s _yobtain bearing capacity p/ s _y:

(8)

Or (8a)

Formula (8) shows, dtlarger, bearing capacity or allowable load p ₂/ s _yless.

Order dt ₁= dt ₂? p/ s _y=ln k. p/ s _y<ln ktime, dt ₁< dt ₂; p/ s _y>ln ktime, dt ₁> dt ₂, this situation engineering generally there will not be, namely generally always dt ₁≤ dt ₂. dt= dt _ctime, formula (7) becomes

or (9)

dt= dt _ctime, formula (8) then becomes

(10)

Order dt ₁≤ dt _c? ; Order dt ₂≤ dt _cresult be p>= p ₄.

As long as dt ₁< dt ₂if, namely p/ s _y<ln kno matter, dt= dt ₁, or dt= dt ₂, all can ensure simultaneously s _to- s _ro≤ s _y, s _ti- s _ri≤ s _y. p ₁= p ₂time,

(11)

Or , n=ln k(11a)

dt _xbe called the desirable temperature difference, kit is larger, dt _xlarger. dt≤ dt _x， p ₁≤ p ₂； dt≥ dt _x， p ₁≥ p ₂。

dt= dt _xtime, formula (7), (8) become

p/ s _y≤ ln kor (12)

Order dt ₁≤ dt _x? p/ s _y≤ ln k; Order dt ₂≤ dt _x? p/ s _y>=ln k. dt= dt _xtime, p ₁/ s _y= p ₂/ s _y=ln k.

Order dt _x= dt _cor make formula (9) the right=formula (10) the right namely p ₃= p ₄all:

(13)

The solution of formula (13) is k= k _c=2.2184574899167 ..., k _cbe called critical footpath ratio. n, idirectly reflect respectively dt _x, dt _csize, n, ias shown in Figure of description 4. k< k _ctime, the inside and outside wall of pressurized container constructed at the utility model maintains dt= dt _xthe temperature difference, its bearing capacity is maximum, for p/ s _y=ln k. k> k _ctime, make dt≤ dt _c; dt= dt _ctime, bearing capacity is maximum, for .In the interior compression elasticity temperature difference prestressing force that visible the utility model constructs, heating self-reinforcing pressure vessel has best bearing capacity or minimum wall thickness (MINI W.).If dt= dt _c, k≤ k _ctime, p ₃>= p ₄; k>= k _ctime, p ₃≤ p ₄.

Formula (12) the right=formula (9) the right is made to obtain:

(14)

The solution of formula (14) is p/ s _y=0.796812 ...= p _c/ s _y, p< p _cor p/ s _y< p _c/ s _ytime, determine footpath ratio with formula (12); p> p _cor p/ s _y> p _c/ s _ytime, determine footpath ratio with formula (9).

Embodiment 1, and somewhere needs a footpath ratio kthe pressurized container of=2, e, a, uas described in example (lower same), s _y=300MPa, what kind of technical characterstic does the pressurized container that construct have?

This is k< k _csituation, if do not make Self-enhancement treatment, its bearing capacity of such container is only p/ s _y= p _e/ s _y=0.375.

1. then apply at the inside and outside wall of pressurized container according to the utility model dt= dt _xthe temperature difference best results of=124.411 ° of C. dt= dt _xtime, p/ s _y=0.693147391(=ln k=ln2>0.375).

If 2. get dt=130 ° of C> dt _x, then p ₁/ s _y=0.707439>ln2, p ₂/ s _y=0.656900602<ln2. p= p ₁time, s _ti- s _ri=300MPa, s _to- s _ro=310MPa> s _y; p= p ₂time, s _ti- s _ri=260MPa, s _to- s _ro=300MPa.Therefore dtduring=130 ° of C, corresponding bearing capacity is p/ s _y=0.656900602<ln2.

If 3. get dt=110 ° of C< dt _x, then p ₁/ s _y=0.656295, p ₂/ s _y=0.786608202. p= p ₁time, s _ti- s _ri=300MPa, s _to- s _ro=274MPa; p= p ₂time, s _ti- s _ri=404MPa> s _y, s _to- s _ro=300MPa.Therefore dtduring=110 ° of C, corresponding bearing capacity is p/ s _y=0.656295<ln2.

4. will dt= dt _c=146.6433 ° of C substitute into formula (7) or incite somebody to action k=2 substitute into formula (9) obtains p/ s _y=0.75>ln k, but p/ s _ywhen=0.75, s _ti- s _ri=300MPa, s _to- s _ro=340MPa> s _y.Will dt= dt _c=146.6433 ° of C substitute into formula (8) or incite somebody to action k=2 substitute into formula (10) obtains p/ s _y=0.548962477<ln k.

So the construction of pressure vessel feature of best results is k=2, dt= dt _x=124.411 ° of C, p/ s _y=0.693147391=ln k, the total stress under this structure characteristic is shown in Figure of description 5.

Embodiment 2, and somewhere needs a footpath ratio kthe pressurized container of=2.5, material as described in example, s _y=300MPa, what kind of technical characterstic should the pressurized container that construct have?

This is k> k _csituation, if do not make Self-enhancement treatment, its bearing capacity of such container is only p/ s _y= p _e/ s _y=0.42.

1. utilize technical measures of the present utility model, apply at the inside and outside wall of pressurized container dt= dt _cthe temperature difference best results of=139.181 ° of C. dt= dt _ctime, corresponding p/ s _y=0.84=2 × 0.42.

If 2. get dt=130 ° of C< dt _c, then p ₁/ s _y=0.812295, p ₂/ s _y=1.274343849. p= p ₁time, s _ti- s _ri=300MPa, s _to- s _ro=247.2MPa; p/ s _y= p ₂time, s _ti- s _ri=630MPa> s _y, s _to- s _ro=300MPa.Therefore dtduring=130 ° of C, corresponding bearing capacity is p/ s _y=0.812295<0.84.

So, for kthe pressurized container of=2.5, within it, outer wall apply dt= dt _ctemperature difference best results.

If the temperature difference can be less than dt _cscope in select, then to a sizing ratio kcylinder, modus ponens (7), (8) determined little person are as allowable load; Or for a fixed difference difference and carrying task ( p/ s _y) cylinder, utilize formula (7), (8) or formula (9), (12) to determine footpath ratio k, get the footpath ratio of large person as constructed pressurized container, and its wall thickness b= r _o- r _i= r _i( k-1), wherein r _idetermined by pressurized container process conditions.Utilize formula (7), (8), can at footpath ratio k, bearing capacity p/ s _y, material property factor ( e, a, u, s _y), the temperature difference dtintercropping parameter adjustment, balance and coupling, heat self-reinforcing pressure vessel structure (see embodiment 3) to construct to meet in industry needs in compression elasticity temperature difference prestressing force.

Embodiment 3, and somewhere needs a pressurized container bearing 225MPa, should have what kind of technical characteristics and just can meet the demands.

If adopt s _ythe material of=350MPa, if do not make Self-enhancement treatment, due under elastic state, when container wall thickness is infinitely great, also can only bear p/ s _y=1/3 ^0.5the pressure of=0.577, and now p/ s _y=225/350=0.642857>0.577, even if so container wall thickness infinity can not meet the demands when not making Self-enhancement treatment.

If 1. need wall thickness minimum, modern p/ s _y<0.796812, then utilize formula (12) determine k=1.901907< k _cas actual footpath ratio.Checking: kwhen=1.901907, dt _x=134.6154 ° of C, dt _c=173.259 ° of C, get dt= dt _x, then s _ti- s _ri= s _to- s _ro=350MPa= s _y.If modus ponens (9) , then dt _x=107.802 ° of C, dt _c=179.1832 ° of C, get dt= dt _x, then s _ti- s _ri=489.4294MPa> s _y, s _to- s _ro=399.7961MPa> s _y.The construction of pressure vessel met the demands is: k=1.901907, dt=134.6154 ° of C.

If do not need wall thickness minimum, then can select under prompting 1. one suitable kvalue,

2. as k=2.2, then dt _x=165.1043 ° of C, dt _c=167.1825 ° of C, dt ₁=103.7427 ° of C, dt ₂=186.349 ° of C> dt _c(meaningless). p/ s _y=0.642857, dt= dt ₁time, s _ti- s _ri=350MPa= s _y, s _to- s _ro=246.8MPa< s _y; p/ s _y=0.642857, dt= dt _ctime, s _ti- s _ri=217.2MPa< s _y, s _to- s _ro=326.1MPa< s _y.If p/ s _y=0.642857, dt= dt ₂, then s _ti- s _ri=177.0618MPa< s _y, s _to- s _ro=350MPa= s _y.So, as long as dt< dt ₂although, dt ₂> dt _c, also have s _ti- s _ri≤ s _y, s _to- s _ro≤ s _y, but dt ₂> dt _cmeaningless, because now container is surrendered already, elasticity temperature difference stress is not formula (1).The construction of pressure vessel met the demands is: k=2.2, dt=103.7427 ° of C; Or k=2.2, dt=167.1825 ° of C.

3. and for example k=3, then dt _c=156.2978 ° of C, dt _x=230.0513 ° of C> dt _c(meaningless), dt ₁=69.77582 ° of C, dt ₂=266.1882 ° of C> dt _c(meaningless). p/ s _y=0.642857, dt= dt ₁time, s _ti- s _ri=350MPa= s _y, s _to- s _ro=133.3MPa< s _y; p/ s _y=0.642857, dt= dt _ctime, s _ti- s _ri=156.3MPa< s _y, s _to- s _ro=228.7MPa< s _y.The construction of pressure vessel met the demands is: k=3, dt=69.77582 ° of C; Or k=3, dt=156.2978 ° of C.

4. obvious, change material behavior ( s _y, e, u, adeng), by different for acquisition p, k, dtcombination, heats self-reinforcing pressure vessel to be formed in different structure feature in compression elasticity temperature difference prestressing force.

Embodiment 4, and certain production process can provide dtthe temperature difference of=120 ° of C, desirable to provide the construction of pressure vessel meeting need of production, if adopt s _ythe material of=350MPa.

(1) if without pressure-bearing restriction, then multiple technologies scheme can be had, as 1. got k=2.5, then p ₁/ s _y=0.785673, p ₂/ s _y=2.48657.Get p= p ₁=0.785673 × 350=275MPa.2. get k=1.6, then p ₁/ s _y=0.506252, p ₂/ s _y=0.402030791.Get p= p ₂=0.402030791 × 350=140.7MPa.And so on.

(2) suppose above-mentioned dt=120 ° of C= dt _x, then k=1.773688, p/ s _y=ln k=0.573060794.

(3) suppose above-mentioned dt=120 ° of C= dt _c, then k=49.872.Unrealistic, this scheme is cast out.

(4) if require pressure-bearing 1. 200MPa, namely p/ s _y=200/350=0.571429, then k=1.76884, or k=1.772141, get large person k=1.772141.2. pressure-bearing 300MPa is required, namely p/ s _y=300/350=0.857143, then k=4.219641, or k=2.019384, get large person k=4.219641.Checking: 1. p/ s _y=0.571429, k=1.772141, dtduring=120 ° of C, s _ti- s _ri=348.9MPa< s _y, s _to- s _ro=350MPa= s _y.2. p/ s _y=0.857143, k=4.219641, dtduring=120 ° of C, s _ti- s _ri=350MPa= s _y, s _to- s _ro=151.1MPa< s _y.

k≤ k _ctime, if dt= dt _x, then p= p _y= s _yln k, and thermal force is also just full yield load p _t= s _yln k= p _y= p.Therefore k≤ k _c, dt= dt _xtime, three-dimensional total stress is

(15)

From formula (15), k≤ k _ctime ,- s _y≤ s _r, s _t, s _z,≤ s _y, and in whole container wall s _t- s _r≡ s _y.This also can the Figure of description 5 of See Examples 1.

k>= k _ctime, if dt= dt _c, then p/ s _y=( k ²-1) k ², thermal force and formula (3).Therefore k>= k _c, dt= dt _ctime, total stress is

(16)

, it is known by analysis, k>= k _ctime, in whole container wall s _t- s _r≤ s _y; x=1, i.e. internal face place, s _ti- s _ri= s _y; x= k, i.e. outer wall place, , k= k _ctime, , therefore, ( s _to- s _ro)/ s _y=1.

Known by formula (7), as long as dt>=0, namely, heating, just has p/ s _y> ( k ²-1)/(2 k ²)= p _e/ s _y, namely p> p _e; Known by formula (8), reduce the temperature difference and can improve bearing capacity p/ s _y, dtwhen=0,

(17)

Make again p ₂/ s _y>=( k ²-1)/(2 k ²) obtain effective temperature difference dt _e

， (18)

Known by formula (18), (11), when time (must set up), no matter namely kwhy be worth, always dt _e>= dt _x. dt= dt _etime, p ₂= p _e; dt< dt _etime, p ₂> p _e, the temperature difference is down to dt _xtime, interior compression elasticity temperature difference prestressing force interior heating self-reinforcing pressure vessel p/ s _yreach full yield load p _y/ s _y.Will k~ zrelation curve adds Fig. 4 to, to compare i, n, z.

Order dt _e= dt _cor order z= ior order p ₄= p _e:

k ⁴ln k ²=( k ²-1)(2 k ²-1) (19)

The solution of formula (19) is k=1.778365= k _e. k _ebe called effective diameter ratio, accordingly p _e/ s _y=0.341901. k< k _etime, z> i, dt _e> dt _c; k> k _etime, z< i, dt _e< dt _c. k< k _econtainer, dtunsuitable excessive, if dt> dt _e, its bearing capacity is less than unexpectedly p _e.

Embodiment 5. kthe technological scheme of the pressurized container of=1.6, s _y=350MPa.

kwhen=1.6, p _e/ s _y=0.304687656, dt _c=181.3931 ° of C, dt _x=98.41956 ° of C< dt _c, dt _e=150.9051 ° of C< dt _c.

If 1. get dt _x< dt=110 ° of C< dt _e, then p ₁/ s _y=0.489455, p ₂/ s _y=0.433528225; Get p ₁time, s _ti- s _ri=350MPa, s _to- s _ro=375MPa> s _y; Get p ₂time, s _ti- s _ri=286MPa< s _y, s _to- s _ro=350MPa= s _y.Therefore dtduring=110 ° of C, corresponding bearing capacity is p/ s _y=0.433528225=1.42 p _e/ s _y.The technical characteristics of container is: k=1.6, dt=110 ° of C, p/ s _y=0.433528225.

If 2. get dt= dt _e, then p ₁/ s _y=0.558164, p ₂/ s _y=0.304687656.Get p ₁time, s _ti- s _ri=350MPa= s _y, s _to- s _ro=464MPa> s _y; Get p ₂time, s _ti- s _ri=59MPa< s _y, s _to- s _ro=350MPa= s _y.Therefore dt= dt _etime, corresponding bearing capacity is p/ s _y=0.304687656= p _e/ s _y.The technical characteristics of container is: k=1.6, dt= dt _e, p/ s _y=0.304687656.But k=1.6, dtalso can carry when=0 p/ s _y=0.304687656, therefore cast out this scheme.

If 3. get dt _e< dt=160 ° of C< dt _c, then p ₁/ s _y=0.573441, p ₂/ s _y=0.276041055.Get p ₁time, s _ti- s _ri=350MPa= s _y, s _to- s _ro=483MPa> s _y; Get p ₂time, s _ti- s _ri=8.4MPa< s _y, s _to- s _ro=350MPa= s _y.Therefore dt=160 ° of C> dt _etime, corresponding bearing capacity is p/ s _y=0.276041055=0.906 p _e/ s _y.The technical characteristics of container is: k=1.6, dt=160 ° of C, p/ s _y=0.276041055< p _e/ s _y.This scheme is also cast out.

If 4. get dt=90 ° of C< dt _x, then p ₁/ s _y=0.455861, p ₂/ s _y=0.496523094.Get p ₁time, s _ti- s _ri=350MPa= s _y, s _to- s _ro=332MPa< s _y; Get p ₂time, s _ti- s _ri=397MPa> s _y, s _to- s _ro=350MPa= s _y.Therefore dt=90 ° of C< dt _xtime, corresponding bearing capacity is p/ s _y=0.455861, p _e/ s _y< p/ s _y< p _y/ s _y=ln1.6=0.47.The technical characteristics of container is: k=1.6, dt=90 ° of C, p/ s _y=0.455861.

5. best technical characteristics is: k=1.6, dt= dt _x, p/ s _y=ln k=0.47000364.

Can be accepted or rejected different technologies scheme according to desirability in engineering reality.Such as, in container work process, inside and outside wall temperature difference has unrestrictedly, if having, then dtwith regard to the actual temperature difference in extracting container work process; To ask bearing capacity high, then high for criterion with bearing capacity; Etc..

If not requirement p/ s _y=ln kor p=2 p _e, in order to do general process, order p ₁= lp _e?

(20)

(20a)

lrepresent that bearing capacity is p _emultiple, l=0 ~ ∞, but l<1 is meaningless.Order p ₂= lp _e?

(21)

(21a)

lduring <1, dt ₃<0, meaning and external heat, and l<1 represents that bearing capacity is less than mechanical incipient yield pressure, does not add the temperature difference and still can bear p _estress, apply temperature difference bearing capacity reduce on the contrary ( l<1), so l<1 is meaningless. lwhen=2, m= i, dt ₃< dt _c; lduring >2, m> i, dt ₃> dt _c; lwhen=1, w= z, dt ₃= dt ₄; lduring >1, w< z, dt ₃< dt ₄. lit is larger, dt ₃larger, dt ₄less.The meaning of formula (20), (21) is: when bearing capacity is p _e's ltimes time, footpath is than the relation that should meet with the temperature difference.For ease of using and comparing, will k~ m, k~ wrelation curve adds Figure of description 4 to.In figure, dotted line is w, all the other are solid line, and curve 1 ~ 10 represents identical llower corresponding w, mcurve, as curve 7 represents lwhen=1.7 w, mcurve.1: l=0,2: l=0.5,3: l=1( mi.e. transverse axis), 4: l=1.1,5: l=1.3,6: l=1.5,7: l=1.7,8: l=1.9,9: l=2,10: l=2.5.Represented by curve 3 mcurve is actually abscissa.The coordinate of a point is a( k _c, 0.796812), the coordinate of b point is b( k _e, 0.841901).All must in any case dt≤ dt _c, and should be actual according to engineering, select to meet objective fact dt, kcombination, choosing preferably of the temperature difference k~ m, k~ wthe intersection point of curve, makes for this reason dt ₃≤ dt ₄or m≤ w?

or (22)

time, namely k= k ₁time, m= w=ln k= n, namely dt ₃= dt ₄= dt _x, so see from Figure of description 4, k~ m, k~ wthe intersection point of curve just exists k~ non curve. dt ₄= dt _x= dt ₃time, p/ s _y= lp _e/ s _y=ln k. l≤ l ₁or k>= k ₁time, dt ₃≤ dt ₄; l>= l ₁or k≤ k ₁time, dt ₃>= dt ₄; lwhen=2, namely above-mentioned rule becomes: k≤ k _ctime, dt ₄≤ dt _c; k>= k _ctime, dt ₄>= dt _c. p= lp _etime, will dt ₃substitution formula (6) also makes s _to- s _ro≤ s _y, will dt ₄substitution formula (5) also makes s _ti- s _ri≤ s _y, result is all formula (22).This means, under meeting the condition of formula (22), can ensure simultaneously s _to- s _ro≤ s _y, s _ti- s _ri≤ s _y; And only have dt ₃≤ dt ₄, just have s _to- s _ro≤ s _y, s _ti- s _ri≤ s _y. lduring >2, dt ₃> dt _c, due to palpus dt ₃≤ dt ₄, so lduring >2, no matter dt ₄size how, all nonsensical in engineering.Another significance of formula (22) meets when being and seeking common ground s _to- s _ro≤ s _y, s _ti- s _ri≤ s _ycondition minimum k, or meet simultaneously s _to- s _ro= s _y, s _ti- s _ri= s _yconditioned disjunction m= wcondition k(be designated as k ₁).As, lduring <1,1≤ kwithin the scope of≤∞, do not exist and make dt ₃= dt ₄'s k ₁; l=1: k ₁=1, m= w=ln k= n=0; l=1.1: k ₁=1.101721, m= w=ln k= n=0.096874; l=1.3: k ₁=1.316439, m= w=ln k= n=0.27493; l=1.5: k ₁=1.548225, m= w=ln k= n=0.437109; l=1.7: k ₁=1.799499, m= w=ln k= n=0.587508; l=1.9: k ₁=2.07276, m= w=ln k= n=0.728881; l=2: k ₁= k _c, m= w=ln k= n=0.796812= i; l=2.5: k ₁=3.052027, m= w=ln k= n=1.115806. k< k ₁or dt ₃>= dt ₄time, without meeting p= lp _e's k, namely without meeting p= lp _etechnological scheme, must have k>= k ₁or dt ₃≤ dt ₄. k>= k ₁time, due to dt ₃≤ dt ₄if, therefore dt ₄≤ dt _c, just have satisfied p= lp _e's k, namely have satisfied p= lp _etechnological scheme.Order dt ₄≤ dt _cor w≤ i?

or (23)

l≤ l ₂or k>= k ₂time, dt ₄>= dt _c; l>= l ₂or k≤ k ₂time, dt ₄≤ dt _c. l≤2， dt ₃≤ dt _c。Simultaneous formula (22), (23) are analyzed: k< k _ctime, l ₂< l ₁<2, k ₂> k ₁; k= k _ctime, l ₂= l ₁=2, k ₂= k ₁= k _c; k> k _ctime, l ₂> l ₁>2, k ₂< k ₁.

k ₁for curve wwith mabscissa (the curve of intersection point wwith mintersection point drop on curve mon); k ₂for curve wwith ithe abscissa of intersection point. l=1： k ₁=1， k ₂=1.778365= k _e、 l=1.1： k ₁=1.101721， k ₂=1.829547、 l=1.3： k ₁=1.316439， k ₂=1.925965、 l=1.5： k ₁=1.548225， k ₂=2.015782、 l=1.6： k ₁=1.671272， k ₂=2.058598、 l=1.683803： k ₁= k _e， k ₂=2.093525、 l=1.7： k ₁=1.799499， k ₂=2.10018、 l=1.9： k ₁=2.07276， k ₂=2.180028、 l=1.95： k ₁=2.144818， k ₂=2.19936、 l=2： k ₁= k ₂= k _c。

k= k ₁time, (when l<2), (when l<2, l< k ₁ ²).

Order dt ₁>= dt ₃? ; Order dt ₂>= dt ₄? . p= lp _etime, dt ₁= dt ₃, dt ₂= dt ₄.

p= lp _e, dt= dt _ctime, s _ti- s _ri=( l-1) s _y, only with lrelevant, and with kirrelevant.

p= lp _etime, formula (10) becomes , namely when dt= dt _ctime, if l≤ l ₂, must have s _to- s _ro≤ s _y.And if l≤ l ₂, then k>= k ₂, so, when dt= dt _ctime, if l≤ l ₂, or k>= k ₂, must have s _to- s _ro≤ s _y.

After vague generalization process, heating self-reinforcing pressure vessel technological scheme in interior compression elasticity temperature difference prestressing force can be constructed more easily.

Embodiment 6.Suppose requirement p=0.5 p _e, material s _y=350MPa, e, a, u, same to example.

1. as k=1.7, then dt _c=178.4203 ° of C, dt _x=111.1145 ° of C< dt _c, dt _e=165.7205 ° of C< dt _c, dt ₃=-89.2102 ° of C(external heats), dt ₄=209.5619 ° of C> dt _c(∵ l=0.5< l ₂=0.855), give up. dt= dt ₃time, p ₁/ s _y=0.163495=0.5 p _e/ s _y, p ₂/ s _y=1.277685644, get p ₁/ s _y, then s _ti- s _ri=350MPa, s _to- s _ro=-63MPa< s _y.

2. as k=2.7, then dt _c=159.7059 ° of C, dt _x=207.9886 ° of C> dt _c(because k> k _c), dt _e=262.2961 ° of C< dt _c, dt ₃=-79.8529 ° of C(external heats), dt ₄=283.1464 ° of C> dt _c(∵ l=0.5< l ₂=3.5), give up. dtduring=-79.8529 ° of C, p ₁/ s _y=0.215707=0.5 p _e/ s _y, p ₂/ s _y=3.971118932, get p ₁, then s _ti- s _ri=350MPa, s _to- s _ro=-68MPa< s _y.

Do not add the temperature difference still can bear p _epressure, apply temperature difference bearing capacity and reduce on the contrary, therefore, the present embodiment shows l<1 does not have practical significance.

Embodiment 7.Suppose requirement p= p _e, material is with embodiment 6.

l=1, then dt=0, appoint and get one k,

1. as k=1.5, then dt _c=184.707 ° of C, dt _x=84.90509 ° of C< dt _c, dt _e=134.2881 ° of C< dt _c, dt ₄=134.2881 ° of C< dt _c(∵ l=1> l ₂=0.53). dtwhen=0, p ₁/ s _y=0.277778= p _e/ s _y, p ₂/ s _y=0.625, get p ₁, then s _ti- s _ri=350MPa, s _to- s _ro=156MPa< s _y.The technical characteristics of container is: k=1.5, dt=0, p=97.2223MPa= p _e. dt= dt _eduring=134.2881 ° of C, p ₁/ s _y=0.479731, p ₂/ s _y=0.277778, get p ₂, then s _ti- s _ri=96MPa< s _y, s _to- s _ro=350MPa= s _y.The technical characteristics of container is: k=1.5, dt=134.2881 ° of C, p=97.2223MPa= p _e.Certainly, last technological scheme is because of better without the need to the temperature difference, and namely one without the temperature difference kthe container of=1.5 can bear p= p _epressure loading.

2. as k=3, then dt _c=156.2978 ° of C, dt _x=230.0513 ° of C> dt _c(because k> k _c), dt _e=281.9203 ° of C> dt _c, dt ₄=281.9203 ° of C= dt _e(∵ l=1< l ₂=4.6), give up.Now one without the temperature difference kthe container of=3 can bear p=155.5556MPa= p _epressure.

Embodiment 8.Suppose requirement p=1.5 p _e, material is with embodiment 6.

1. as k=1.4, then dt _c=188.4261 ° of C, dt _x=70.45786 ° of C< dt _c, dt _e=115.4117 ° of C< dt _c, dt ₃=94.21306 ° of C, dt ₄=55.30145 ° of C< dt _c. dt= dt ₃time, p ₁/ s _y=0.367346944=1.5 p _e/ s _y, p ₂/ s _y=0.288081178=1.176 p _e/ s _y.Get p ₁, then s _ti- s _ri=350MPa, s _to- s _ro=408MPa> s _y.Get p ₂, then s _ti- s _ri=234MPa, s _to- s _ro=350MPa= s _y, but now bearing capacity is less than required p=1.5 p _e. dt= dt ₄time, p ₁/ s _y=0.316773=1.3 p _e/ s _y, p ₂/ s _y=0.367346944=1.5 p _e/ s _y.Get p ₁, then s _ti- s _ri=350MPa, s _to- s _ro=313MPa, but now bearing capacity is less than required p=1.5 p _e, give up.Due to l(1.5) > l ₁(1.37) or k(1.4) < k ₁(1.548225) time, so, get p ₂/ s _ymust make s _ti- s _ri> s _y.Therefore kwithout suitable technological scheme when=1.4.

Meet when utilizing formula (23) to seek common ground below s _to- s _ro≤ s _y, s _ti- s _ri≤ s _ymost path ratio k= k ₁. lwhen=1.5, k ₁=1.548225. k= k ₁time, dt _c=183.0618 ° of C, dt _x=91.53478 ° of C< dt _c, dt _e=142.5564 ° of C< dt _c, dt ₃= dt ₄= dt _x=91.53478 ° of C, so p ₁/ s _y= p ₂/ s _y=0.437125252=1.5 p _e/ s _y, s _ti- s _ri= s _to- s _ro=350MPa.The technical characteristics of container is: k=1.548225, dt=91.53478 ° of C, p=153MPa=1.5 p _e.Formula (24) is utilized to try to achieve k ₂=2.015782.

2. as k=1.65, then dt _c=179.8676 ° of C, dt _x=104.8632 ° of C< dt _c, dt _e=158.5145 ° of C< dt _c, dt ₃=89.93378 ° of C, dt ₃< dt ₄=112.5016 ° of C< dt _c(∵ l=1.5> l ₂=0.77). dt= dt ₃time, p ₁/ s _y=0.474517749=1.5 p _e/ s _y, p ₂/ s _y=0.552096235> p ₁/ s _y(giving up), s _ti- s _ri=350MPa, s _to- s _ro=318.5MPa< s _y.The technical characteristics of container is: k=1.65, dt=89.93378 ° of C, p= p ₁=166.08121MPa=1.5 p _e. dt= dt ₄time, p ₁/ s _y=0.514209>1.5 p _e/ s _y(giving up), p ₂/ s _y=0.474517749=1.5 p _e/ s _y< p ₁/ s _y, s _ti- s _ri=306.1MPa< s _y, s _to- s _ro=350MPa= s _y.The technical characteristics of container is: k=1.65, dt=112.5016 ° of C, p= p ₂=166.08121MPa=1.5 p _e.Certainly, if wish, the temperature difference is too not high, then last technological scheme is better.

3. as k=2.4, then dt _c=163.861 ° of C, dt _x=183.3247 ° of C> dt _c, dt _e=239.652 ° of C> dt _c, dt ₃=81.9305, dt ₄=214.4785 ° of C> dt _c. dt= dt ₃time, p ₁/ s _y=0.619792=1.5 p _e/ s _y, p ₂/ s _y=1.707602706(gives up), s _ti- s _ri=350MPa, s _to- s _ro=190MPa.The technical characteristics of container is: k=2.4, dt= dt ₃, p= p ₁=216.9172MPa=1.5 p _e.

Embodiment 9.Suppose requirement p=2 p _e, material is with embodiment 6.

Meet when 1. first utilizing formula (22) to seek common ground s _to- s _ro≤ s _y, s _ti- s _ri≤ s _y's k ₁. lwhen=2, formula (22) is formula (13), so k ₁= k _c=2.2184574899167. k= k _ctime, dt _c= dt _x=166.8538 ° of C, dt _e=223.9652 ° of C> dt _c, dt ₃= dt ₄= dt _x= dt _c=166.8538 ° of C, p ₁/ s _y= p ₂/ s _y=0.796812284=2 p _e/ s _y, s _ti- s _ri= s _to- s _ro=350MPa.So the technical characteristics of container is: k= k _c, dt=166.8538 ° of C, p=278.8843MPa=2 p _e.

2. as got k=2.8, then dt _c=158.4978 ° of C, dt _x=215.6041 ° of C> dt _c, dt _e=269.1262 ° of C> dt _c, dt ₃=158.4978 ° of C, dt ₄=229.7803 ° of C> dt _c. dt= dt ₃time, p ₁/ s _y=0.872449=2 p _e/ s _y, p ₂/ s _y=1.662750841(gives up), s _ti- s _ri=350MPa, s _to- s _ro=270MPa.So the technical characteristics of container is: k=2.8, dt= dt ₃, p=305.35715MPa=2 p _e.

Since k< k ₁or dt ₃>= dt ₄time, without meeting p= lp _etechnological scheme, l<1 is meaningless, ithe above region of curve is also meaningless, therefore by curve m, wbe modified to Figure of description 6.(consult Figure of description 6) in sum:

(1) to arbitrary l, k ₁≤ k≤ k ₂in scope, to arbitrary k, have 3 can ensure simultaneously s _ti- s _ri≤ s _y, s _to- s _ro≤ s _ythe temperature difference: dt ₃( dt ₁)≤ dt _x≤ dt ₄( dt ₂). k= k ₁time, dt ₃= dt ₄; k> k ₁time, dt ₃< dt ₄; And dt ₃= dt ₄time, dt ₃= dt ₄= dt _x.So, k ₁≤ k≤ k ₂in scope, dt ₃≤ dt _x≤ dt ₄. lwhen≤2, k ₂≤ k _c, and k ₂≤ k _ctime, dt ₄≤ dt _c.But dt= dt _xtime, p/ s _y=ln k≠ lp _e/ s _y, so, only have 2 to ensure simultaneously s _ti- s _ri≤ s _y, s _to- s _ro≤ s _y, p= lp _ethe temperature difference: dt ₃( dt ₁)≤ dt ₄( dt ₂).

(2) to arbitrary l, k ₂≤ k≤ k _cin scope, to arbitrary k, have 3 can ensure simultaneously s _ti- s _ri≤ s _y, s _to- s _ro≤ s _ythe temperature difference: dt ₃( dt ₁)≤ dt _x≤ dt _c. k> k ₂time, dt ₄> dt _c, invalid. lwhen≤2, dt ₃≤ dt _c; k≤ k _ctime, dt _x≤ dt _c. l≤ l ₁time, dt ₃≤ dt _x. kincrease, l ₁increase, k= k ₁time, l= l _1min, so, k>= k ₁time, l≤ l ₁. p= lp _e, dt= dt _ctime, s _ti- s _ri=( l-1) s _y< s _y( l≤ 2); p= lp _e, k>= k ₂, dt= dt _ctime, s _to- s _ro≤ s _y.But dt= dt _xtime, p/ s _y=ln k≠ lp _e/ s _y; dt= dt _ctime, p= p ₃or p= p ₄, p≠ lp _e.So, only have 1 to ensure simultaneously s _ti- s _ri≤ s _y, s _to- s _ro≤ s _y, p= lp _ethe temperature difference: dt ₃( dt ₁).

(3) to arbitrary l, k _c≤ kwithin the scope of≤∞, to arbitrary k, have 2 can ensure simultaneously s _ti- s _ri≤ s _y, s _to- s _ro≤ s _ythe temperature difference: dt ₃( dt ₁)≤ dt _c.As long as l≤ 2, namely have dt ₃≤ dt _c. p= lp _e, dt= dt _ctime, s _ti- s _ri≤ s _y, s _to- s _ro≤ s _y.But dt= dt _ctime, p≠ lp _e.So, only have 1 to ensure simultaneously s _ti- s _ri≤ s _y, s _to- s _ro≤ s _y, p= lp _ethe temperature difference: dt ₃( dt ₁).

Embodiment 10.Suppose requirement p=1.6 p _e, material with embodiment 6, s _y=350MPa, formulates the technological scheme met the demands.

Technological scheme compares in table 1.

Table 1 p=1.6 p _e, Different Diameter than under technological scheme

Continued 1

Figure of description 7 shows k ₁=1.671272, lwhen=1.6 m, wcurve and i, z, ncurve.

Embodiment 11.Suppose requirement p=1.9 p _e, material with embodiment 6, s _y=350MPa, formulates the technological scheme met the demands.

lwhen=1.9, make respectively k ₁=2.07276, k ₂=2.180028.According to above analysis,

(1) exist k ₁≤ k≤ k ₂in scope, there are 2 to ensure simultaneously s _ti- s _ri≤ s _y, s _to- s _ro≤ s _y, p= lp _ethe temperature difference: dt ₃( dt ₁)≤ dt ₄( dt ₂).Get k=2.1, then dt ₃=152.1461 ° of C, dt ₄=156.5498 ° of C.The technical characteristics of container is: k=2.1, dt= dt ₃, p=1.9 p _e=257.1032Mpa; Or k=2.1, dt= dt ₄, p=1.9 p _e=257.1032Mpa.

(2) exist k ₂≤ k≤ k _cin scope, there is 1 to ensure simultaneously s _ti- s _ri≤ s _y, s _to- s _ro≤ s _y, p= lp _ethe temperature difference: dt ₃( dt ₁).Get k=2.2, then dt ₃=150.4642 ° of C.The technical characteristics of container is: k=2.2, dt= dt ₃, p=1.9 p _e=263.8017Mpa.

(3) exist k _c≤ kwithin the scope of≤∞, there is 1 to ensure simultaneously s _ti- s _ri≤ s _y, s _to- s _ro≤ s _y, p= lp _ethe temperature difference: dt ₃( dt ₁).Get k=2.3, then dt ₃=148.9121 ° of C.The technical characteristics of container is: k=2.3, dt= dt ₃, p=1.9 p _e=269.6456Mpa.If get k=3.5, then dt ₃=136.6117 ° of C.The technical characteristics of container is: k=3.5, dt= dt ₃, p=1.9 p _e=305.3571Mpa.……

Claims (2)

1. heating self-reinforcing pressure vessel in compression elasticity temperature difference prestressing force in a kind, it is characterized in that: the inside and outside wall of described pressure vessel exists the temperature difference, the temperature difference makes pressure vessel wall produce temperature difference prestressing force, total stress when described pressure vessel works is that temperature difference prestressing force and mechanical stress are superimposed, described mechanical stress finger pressure container work pressure pThe stress produced, described pressure vessel operating pressure pFor interior pressure; Described internal pressure vessel walls radius surface is r _i, outside wall surface radius be r _o, internal face temperature is t _i, outside wall surface temperature be t _o, the temperature difference Dt= t _i- t _o, t _i> t _o, it is called interior heating; Described pressure vessel, under temperature difference prestressing force effect, is in elastic stage, namely Dt≤ Dt _c, Dt _cFor temperature difference prestressing force has just made internal pressure vessel walls face produce the temperature difference of surrender, it is called critical temperature difference, ,Wherein EFor container material elastic modelling quantity, aFor container material thermal coefficient of expansion, uFor container material Poisson's ratio, s _yFor container material yield strength, kFor the footpath ratio of container, k= r _o/ r _i; k>= k _cTime, its temperature difference is defined as Dt≤ Dt _c, bearing capacity p/ s _yWith Regulation; Dt= Dt _cTime, bearing capacity is ,Now container footpath ratio is ; Wherein p _eFor initial yield mechanical load or maximum machine elastic load, k _cFor by formula Determine kValue, is called critical footpath ratio, k _c=2.2184574899167 k≤ k _cTime, its temperature difference is defined as Dt≤ Dt _x, bearing capacity p/ s _yWith Regulation; Dt= Dt _xTime, bearing capacity is p= s _yLn k= p _y, now container footpath ratio is ; Wherein , be called the desirable temperature difference, p _yFor entirely surrendering mechanical load or integral yield mechanical load; p≤ p _cOr p/ s _y≤ p _c/ s _yTime,Its footpath ratio is ; p>= p _cOr p/ s _y>= p _c/ s _yTime, its footpath ratio is ; Wherein p _cFor by formula Determine pValue, p _c/ s _y=0.796812 The physical dimension of this pressure vessel,I.e. footpath ratio k, bearing capacity p/ s _y, material property factor ( E, a, u, s _y), the temperature difference DtBetween with Or Relation constraint, container wall thickness b= r _o- r _i= r _i( k-1); k, p, E, a, u, s _y, DtSize adjust according to actual needs, combine, mate.

2. heating self-reinforcing pressure vessel in interior compression elasticity temperature difference prestressing force according to claim 1, is characterized in that: the described temperature difference is caused by the operating temperature of pressurized container itself, or heats the inside and outside wall of pressurized container or cool and cause.