CN102592024A - Heat-network modeling method for determining maximum value of steady-state temperature of heat conduction in radial direction - Google Patents

Abstract

The invention provides a heat-network modeling method for determining the maximum value of steady-state temperature of heat conduction in a radial direction, which is applied to the heat conduction of a circular wall cylinder structure. The method provided by the invention comprises the following steps: firstly, using a network with the maximum value of the steady-state temperature of the heat condition in the radial direction to establish a heat-network model of a research object, determining the temperatures of all nodes in the established heat-network model through a heat balance equation, then according to the temperatures of the inner surface and the outer surface of each circular wall cylinder, determining the distribution of the maximum value point rmax of the steady-state temperature of the heat conduction of each circular wall cylinder in the radial direction, and finally, according to the distribution of the maximum value point rmax of the temperature, designing an error compensation link, finally, obtaining the maximum value of the steady-state temperature of the heat conduction of each circular wall cylinder in the radial direction.. According to the method of the invention, the maximum value of the steady-state temperature of the heat conduction in the radial direction of the circular wall cylinder can be determined by the heat-network method, the determined maximum value of the temperature is accurate, the calculation speed is high, the modeling is intuitive, and the obtained value can be further used for analyzing the heat conduction problem of electronic devices.

Description

A kind of definite radially heat supply network network modeling method of heat conduction steady temperature maximum value

Technical field

The present invention relates to electronics heat conduction field, be specifically related to a kind of radially heat supply network network modeling method of heat conduction steady temperature maximum value of confirming.

Background technology

Relate in the electron trade at satellite, Aero-Space, machinery etc., heat dissipation problem is the problem that researcher will solve always, solve complicated heat dissipation problem, just needs to analyze the heat conduction problem of various situation.Hot network technique is a kind of method (reference paper 1: Kang Qin, Li Shiwu, Guo Jianli that compares and find the solution heat conduction problem through thermoelectricity; " hot network technique outline ", " industry heating " the 35th volume, 2006 the 5th phases); This method is at first set up the heat supply network network model of institute's research object; And then obtain the thermal balance equation of institute's research object heat supply network network, through finding the solution thermal balance equation, draw the temperature and the rate of change of each network node in institute's research object.Advantages such as it is fast that hot network technique has the speed of finding the solution, and modeling is directly perceived.

When finding the solution the heat conduction problem of some object, when for example utilizing hot network technique to find the solution the distribution of motor steady temperature, need set up the heat supply network network model of motor through hot network technique.The parts of many motors, such as housing, stator end, rotors etc. can be approximately round wall housing structure according to its features of shape and set up heat supply network network model.Therefore, the heat supply network network model of justifying the wall housing structure is for using hot network technique solution heat conduction problem aspect to have great importance.

Heat conducts on circle wall housing structure; Can be along axially; Radially and circumferential three directions carry out, when handling general heat conduction problem, for making problem reduction; Usually ignore the heat conduction of round wall cylinder body on circumferential, circle wall cylinder body heat conduction this moment just becomes axially the two-dimentional heat conduction problem that makes progress with the footpath.

The hot network technique of current use is found the solution round wall Boring mill when the two-dimentional heat conduction problem that makes progress with the footpath, and using more is mellor heat supply network network model.Can set up round wall cylinder body being the heat supply network network model of network node axially through mellor heat supply network network model with stable state medial temperature radially; Therefore adopting the hot network technique of mellor heat supply network network model is a kind of steady temperature average heat supply network network method, and that obtain is medial temperature (reference paper 2:P.H.Mellor, the et al. of circle wall housing structure; " Lumped parameter thermal model for electrical machines of TEFC design; " Electric Power Applications, IEE Proceedings B, vol.138; Pp.205-218,1991).Yet in finding the solution some heat conduction problem; Medial temperature when the emphasis of being concerned about not is stable state; But the temperature maximum during stable state, this moment, the steady temperature mean value of gained can not satisfy the demands if use mellor heat supply network network model to remove to set up the heat supply network network and then find the solution.To this problem, current to have a kind of Gerling heat supply network network model can set up to justify the wall Boring mill be the heat supply network network of node to heat conduction steady temperature maximum value, but only be to axial heat conduction, to circle wall cylinder body radially heat conduction still can't use.

Summary of the invention

The present invention is directed to current use heat supply network network method and can't find the solution the radially problem of heat conduction steady temperature maximum value of round wall cylinder body, a kind of radially heat supply network network modeling method of heat conduction steady temperature maximum value of circle wall cylinder body that is used for confirming is provided.

A kind of definite radially heat supply network network modeling method of heat conduction steady temperature maximum value may further comprise the steps:

Step 1, use heat conduction steady temperature maximum value network radially to set up the heat supply network network model of institute's research object, list thermal balance equation, find the solution neutral temperature and temperature maximum in the heat supply network network model of being set up.

The heat supply network network model of the foundation described in the step 1, specifically: the structure to each circle wall cylinder body is formed is provided with three thermal resistance R ₁, R ₂And R _m, the first thermal resistance R ₁Right-hand member, the second thermal resistance R ₂Left end and the 3rd thermal resistance R _mLeft end all connect together, three thermal resistances are the Y type and connect, the temperature at interface place is neutral temperature T _m, circle wall inner surface of cylinder block temperature T ₁At the first thermal resistance R ₁Left end, circle wall outer surface of cylinder block temperature T ₂At the second thermal resistance R ₂Right-hand member, temperature maximum T _MaxAt the 3rd thermal resistance R _mRight-hand member, the 3rd thermal resistance R _mThe right-hand member inside that connects circle wall cylinder body give birth to heat G; If the structure that research object is made up of plural round wall cylinder body, then its heat supply network network model is made up of the heat supply network network model connection of each circle wall cylinder body.

Described in the step 1 thermal balance equation be: [Y] [T]+[G]=0, wherein, [Y] is the thermal conductance matrix, [T] is temperature node matrix, [G] is the heat matrix, through [T]=-[Y] ^-1[G], the steady temperature value of each temperature node in the heat supply network network model that obtains being set up.

Step 2, according to the heat supply network network model and the resulting neutral temperature T of the research object of being set up _mWith temperature maximum T _Max, can confirm the inside and outside surface temperature of each circle wall cylinder body, confirm radially heat conduction steady temperature maximum point r separately according to each circle wall cylinder body inside and outside surface temperature separately _MaxBe to be distributed on the inside surface or outside surface of round wall cylinder body, still be distributed between the internal diameter and external diameter of round wall cylinder body.。

Step 3, if heat conduction steady temperature maximum point r radially _MaxBe distributed on the inside surface or outside surface of round wall cylinder body, the temperature maximum that is then obtained by step 1 is exactly final heat conduction steady temperature maximum value T ' _Max=T _MaxIf heat conduction steady temperature maximum point r radially _MaxBe distributed between the internal diameter and external diameter of round wall cylinder body, then confirm approximate compensation tache Δ T, confirm final heat conduction steady temperature maximum value T ' then _Max=T _Max+ Δ T.

Advantage of the present invention and good effect are: modeling method of the present invention is through setting up radially heat conduction steady temperature maximum value heat supply network network model; Make and to use hot network technique to go the problem of the radially heat conduction steady temperature maximum value of definite circle wall cylinder body; With the radially heat conduction steady temperature maximum value comparison of the determined round wall cylinder body of employing mellor heat supply network network model, final determined value is more accurate.Advantages such as the computing speed that the inventive method uses heat supply network network method to have is fast, and modeling is directly perceived can be confirmed the radially heat conduction steady temperature maximum value of circle wall cylinder body, and then can be used for the heat conduction problem of analytical electron device.

Description of drawings

Fig. 1 is the radially overall flow figure of the heat supply network network modeling method of heat conduction steady temperature maximum value that confirms of the present invention;

Fig. 2 is the structural representation of the round wall cylinder body object that is directed against of the inventive method;

Fig. 3 is a function $y=\frac{1}{2}\·\frac{1}{\mathrm{Ln}x}(x^2-1)$ The codomain synoptic diagram;

Fig. 4 is the synoptic diagram of the radially heat conduction steady temperature maximum value heat supply network network model built in the step 2 of the present invention;

Fig. 5 is the complete radially heat conduction steady temperature maximum value heat supply network network model synoptic diagram that step 3 kind of the present invention is built;

Fig. 6 is the structural representation of subjects in the embodiment of the invention;

Fig. 7 is the heat supply network network model synoptic diagram of the subjects in the embodiment of the invention;

Fig. 8 is the subjects stator core steady temperature distribution curve synoptic diagram radially in the embodiment of the invention;

Fig. 9 is the subjects stator teeth steady temperature distribution curve synoptic diagram radially in the embodiment of the invention.

Embodiment

To combine accompanying drawing and embodiment that the present invention is done further detailed description below.

As shown in Figure 1, the present invention is a kind of to confirm that radially the step of the heat supply network network modeling method of heat conduction steady temperature maximum value is following:

Step 1, use heat conduction steady temperature maximum value network radially to set up the heat supply network network model of institute's research object, list thermal balance equation, find the solution the temperature of each temperature node.

Be depicted as the structural representation of round wall cylinder body like accompanying drawing 2, L is the axial length of circle wall cylinder body, r ₁Be circle wall cylinder diameter, r ₂Be circle wall cylinder body external diameter.T ₁Be circle wall inner surface of cylinder block temperature, T ₂Be circle wall outer surface of cylinder block temperature.

Radially heat conduction steady-state heat balance equation is shown below:

$\frac{\&PartialD;T^2\left(r\right)}{\&PartialD;{\mathrm{<figure-callout\; id="2"\; label="r"\; filenames="HDA0000129021630000022.png,HDA0000129021630000023.png"\; state="\{\{state\}\}">r</figure-callout>}}^2}+\frac{1}{r}\frac{\&PartialD;T\left(r\right)}{\&PartialD;r}+\frac{\stackrel{\&CenterDot;}{g}}{k}=0---\left(1\right)$

Wherein: T (r) is radially heat conduction Temperature Distribution of circle wall cylinder body function; The parameter r of T (r) representes the radius radially of circle wall cylinder body;

is the heat-conduction coefficient of circle wall cylinder block material for the heat generation rate of circle wall cylinder body unit volume, k.

Formula (1) is the non-secondly differential equation of second order constant coefficient, and its general solution can be expressed as following form:

$T\left(r\right)=-\frac{\stackrel{\&CenterDot;}{g}}{4\mathrm{<figure-callout\; id="2"\; label="k\; r"\; filenames="HDA0000129021630000022.png,HDA0000129021630000023.png"\; state="\{\{state\}\}">k</figure-callout>}}{r}^{}{}^2+{\mathrm{<figure-callout\; id="1"\; label="c"\; filenames="HDA0000129021630000022.png,HDA0000129021630000032.png"\; state="\{\{state\}\}">c</figure-callout>}}_1\ln r+c_2---\left(2\right)$

Wherein: c ₁And c ₂Be constant.To the following formula differentiate, make that derivative is 0, can get following formula stationary point r ₀(derivative is 0 point),

Ask the second derivative of following formula, with stationary point r ₀Substitution can get

Second derivative is located less than 0, so stationary point r in the stationary point ₀Be maximum point, promptly ${\mathrm{<figure-callout\; id="2"\; label="r\; Max"\; filenames="HDA0000129021630000022.png,HDA0000129021630000023.png"\; state="\{\{state\}\}">r</figure-callout>}}_{\mathrm{Max}}^{}={\mathrm{<figure-callout\; id="0"\; label="r"\; filenames="HDA0000129021630000022.png"\; state="\{\{state\}\}">r</figure-callout>}}_0^2=\frac{2k}{\stackrel{\&CenterDot;}{g}}\&CenterDot;{\mathrm{<figure-callout\; id="1"\; label="c"\; filenames="HDA0000129021630000022.png,HDA0000129021630000032.png"\; state="\{\{state\}\}">c</figure-callout>}}_1.$

If boundary condition T is (r ₁)=T ₁, T (r ₂)=T ₂, boundary condition is brought in the general solution shown in the formula (2), can try to achieve constant c ₁And c ₂Expression formula following:

${\mathrm{<figure-callout\; id="1"\; label="c"\; filenames="HDA0000129021630000022.png,HDA0000129021630000032.png"\; state="\{\{state\}\}">c</figure-callout>}}_1=\frac{1}{\ln \frac{r_2}{r_1}}[(T_2-T_1)+\frac{\stackrel{\&CenterDot;}{g}}{4k}({\mathrm{<figure-callout\; id="2"\; label="r"\; filenames="HDA0000129021630000022.png,HDA0000129021630000023.png"\; state="\{\{state\}\}">r</figure-callout>}}_2^2-{\mathrm{<figure-callout\; id="1"\; label="r"\; filenames="HDA0000129021630000022.png,HDA0000129021630000032.png"\; state="\{\{state\}\}">r</figure-callout>}}_1^2)]---\left(3\right)$

${\mathrm{<figure-callout\; id="2"\; label="c"\; filenames="HDA0000129021630000022.png,HDA0000129021630000023.png"\; state="\{\{state\}\}">c</figure-callout>}}_2=\frac{1}{\ln \frac{r_2}{r_1}}(\mathrm{<figure-callout\; id="2"\; label="ln\; r"\; filenames="HDA0000129021630000022.png,HDA0000129021630000023.png"\; state="\{\{state\}\}">ln</figure-callout>}{r}_{}{}_2\&CenterDot;T_1-\mathrm{<figure-callout\; id="1"\; label="ln\; r"\; filenames="HDA0000129021630000022.png,HDA0000129021630000032.png"\; state="\{\{state\}\}">ln</figure-callout>}{r}_{}{}_1\&CenterDot;T_2)+\frac{\stackrel{\&CenterDot;}{g}}{4k}\&CenterDot;\frac{1}{\ln \frac{r_2}{r_1}}({\mathrm{<figure-callout\; id="2"\; label="ln\; r"\; filenames="HDA0000129021630000022.png,HDA0000129021630000023.png"\; state="\{\{state\}\}">ln</figure-callout>}r}_2\&CenterDot;{\mathrm{<figure-callout\; id="1"\; label="r"\; filenames="HDA0000129021630000022.png,HDA0000129021630000032.png"\; state="\{\{state\}\}">r</figure-callout>}}_1^2-\mathrm{<figure-callout\; id="1"\; label="ln\; r"\; filenames="HDA0000129021630000022.png,HDA0000129021630000032.png"\; state="\{\{state\}\}">ln</figure-callout>}{r}_{}{}_1\&CenterDot;{\mathrm{<figure-callout\; id="2"\; label="r"\; filenames="HDA0000129021630000022.png,HDA0000129021630000023.png"\; state="\{\{state\}\}">r</figure-callout>}}_2^2)---\left(4\right)$

With c ₁Expression formula bring into

Obtain radially heat conduction steady temperature maximum point r _MaxExpression formula be shown below:

$r_{\max }^2=\frac{2k}{\stackrel{\&CenterDot;}{g}}\&CenterDot;\frac{1}{\ln \frac{r_2}{r_1}}[(T_2-T_1)+\frac{\stackrel{\&CenterDot;}{g}}{4k}({\mathrm{<figure-callout\; id="2"\; label="r"\; filenames="HDA0000129021630000022.png,HDA0000129021630000023.png"\; state="\{\{state\}\}">r</figure-callout>}}_2^2-{\mathrm{<figure-callout\; id="1"\; label="r"\; filenames="HDA0000129021630000022.png,HDA0000129021630000032.png"\; state="\{\{state\}\}">r</figure-callout>}}_1^2)]---\left(5\right)$

In the general solution with the radially heat conduction steady-state heat balance equation shown in formula (5) the substitution formula (2), can get temperature maximum T _MaxExpression formula as shown in the formula:

$T_{\max }=-\frac{1}{2}{\mathrm{<figure-callout\; id="1"\; label="c"\; filenames="HDA0000129021630000022.png,HDA0000129021630000032.png"\; state="\{\{state\}\}">c</figure-callout>}}_1+{\mathrm{<figure-callout\; id="1"\; label="c"\; filenames="HDA0000129021630000022.png,HDA0000129021630000032.png"\; state="\{\{state\}\}">c</figure-callout>}}_1\ln {(\frac{2k}{\stackrel{\&CenterDot;}{g}}\&CenterDot;c_1)}^{\frac{1}{2}}+c_2---\left(6\right)$

in the formula (6) item is done linear transformation, it is transformed to:

${\mathrm{<figure-callout\; id="1"\; label="c"\; filenames="HDA0000129021630000022.png,HDA0000129021630000032.png"\; state="\{\{state\}\}">c</figure-callout>}}_1\ln {(\frac{2k}{\stackrel{\&CenterDot;}{g}}\&CenterDot;c_1)}^{\frac{1}{2}}=\frac{\stackrel{\&CenterDot;}{g}}{4k}\&CenterDot;(\frac{2k}{\stackrel{\&CenterDot;}{g}}\&CenterDot;c_1)\&CenterDot;\ln (\frac{2k}{\stackrel{\&CenterDot;}{g}}\&CenterDot;c_1)---\left(7\right)$

Because

can get following formula so make with in the x substitution formula (7):

${\mathrm{<figure-callout\; id="1"\; label="c"\; filenames="HDA0000129021630000022.png,HDA0000129021630000032.png"\; state="\{\{state\}\}">c</figure-callout>}}_1\ln {(\frac{2k}{\stackrel{\&CenterDot;}{g}}\&CenterDot;c_1)}^{\frac{1}{2}}=\frac{\stackrel{\&CenterDot;}{g}}{4k}\&CenterDot;x\&CenterDot;\ln x,x\&Element;[{\mathrm{<figure-callout\; id="1"\; label="r"\; filenames="HDA0000129021630000022.png,HDA0000129021630000032.png"\; state="\{\{state\}\}">r</figure-callout>}}_1^2,{\mathrm{<figure-callout\; id="2"\; label="r"\; filenames="HDA0000129021630000022.png,HDA0000129021630000023.png"\; state="\{\{state\}\}">r</figure-callout>}}_2^2]---\left(8\right)$

Function xlnx is carried out the first-order linear match in

scope, the result after the match is:

x·lnx≈f ₁x+f ₂ (9)

Wherein: f ₁And f ₂Be the coefficient that the first-order linear match obtains, can be to the first-order linear match of function xlnx through the realization of MATLAB function polyfit function, concrete statement is following:

x＝r ₁*r ₁：0.00001：r ₂*r ₂；

y＝x*log(x)；

p＝polyfit(x，y，1)；

f ₁＝p(1，1)；

f ₂＝p(1，2)；

In the linear transform with the substitution as a result after the xlnx match , can get:

${\mathrm{<figure-callout\; id="1"\; label="c"\; filenames="HDA0000129021630000022.png,HDA0000129021630000032.png"\; state="\{\{state\}\}">c</figure-callout>}}_1\ln {(\frac{2k}{\stackrel{\&CenterDot;}{g}}\&CenterDot;c_1)}^{\frac{1}{2}}\&ap;\frac{1}{2}{\mathrm{<figure-callout\; id="1"\; label="f"\; filenames="HDA0000129021630000022.png,HDA0000129021630000032.png"\; state="\{\{state\}\}">f</figure-callout>}}_1\&CenterDot;{\mathrm{<figure-callout\; id="1"\; label="c"\; filenames="HDA0000129021630000022.png,HDA0000129021630000032.png"\; state="\{\{state\}\}">c</figure-callout>}}_1+\frac{\stackrel{\&CenterDot;}{g}}{4k}\&CenterDot;f_2---\left(10\right)$

Just will through above-mentioned linear transformation and first-order linear approximating method

Item is converted into T ₁, T ₂With Linear combination.With fitting result and formula (3) and (4) constant c ₁And c ₂The temperature maximum expression formula T of expression formula substitution formula (6) _MaxIn, can get following formula:

$T_{\max }=\frac{(\frac{1}{2}{f}_1-\frac{1}{2}-\ln r_1)}{\ln \frac{r_2}{r_1}}{\mathrm{<figure-callout\; id="2"\; label="T"\; filenames="HDA0000129021630000022.png,HDA0000129021630000023.png"\; state="\{\{state\}\}">T</figure-callout>}}_2+\frac{(-\frac{1}{2}{\mathrm{<figure-callout\; id="1"\; label="f"\; filenames="HDA0000129021630000022.png,HDA0000129021630000032.png"\; state="\{\{state\}\}">f</figure-callout>}}_1+\frac{1}{2}+\ln r_2)}{\ln \frac{r_2}{r_1}}{\mathrm{<figure-callout\; id="1"\; label="T"\; filenames="HDA0000129021630000022.png,HDA0000129021630000032.png"\; state="\{\{state\}\}">T</figure-callout>}}_1+[\frac{(\frac{1}{2}{f}_1-\frac{1}{2})({\mathrm{<figure-callout\; id="2"\; label="r"\; filenames="HDA0000129021630000022.png,HDA0000129021630000023.png"\; state="\{\{state\}\}">r</figure-callout>}}_2^2-{\mathrm{<figure-callout\; id="1"\; label="r"\; filenames="HDA0000129021630000022.png,HDA0000129021630000032.png"\; state="\{\{state\}\}">r</figure-callout>}}_1^2)+({\mathrm{<figure-callout\; id="2"\; label="ln\; r"\; filenames="HDA0000129021630000022.png,HDA0000129021630000023.png"\; state="\{\{state\}\}">ln</figure-callout>}r}_2\&CenterDot;{\mathrm{<figure-callout\; id="1"\; label="r"\; filenames="HDA0000129021630000022.png,HDA0000129021630000032.png"\; state="\{\{state\}\}">r</figure-callout>}}_1^2-\mathrm{<figure-callout\; id="1"\; label="ln\; r"\; filenames="HDA0000129021630000022.png,HDA0000129021630000032.png"\; state="\{\{state\}\}">ln</figure-callout>}{r}_{}{}_1\&CenterDot;{\mathrm{<figure-callout\; id="2"\; label="r"\; filenames="HDA0000129021630000022.png,HDA0000129021630000023.png"\; state="\{\{state\}\}">r</figure-callout>}}_2^2)}{\ln \frac{r_2}{{\mathrm{<figure-callout\; id="1"\; label="r"\; filenames="HDA0000129021630000022.png,HDA0000129021630000032.png"\; state="\{\{state\}\}">r</figure-callout>}}_1}\&CenterDot;4k}+\frac{{\mathrm{<figure-callout\; id="2"\; label="f"\; filenames="HDA0000129021630000022.png,HDA0000129021630000023.png"\; state="\{\{state\}\}">f</figure-callout>}}_2}{4k}]\&CenterDot;\stackrel{\&CenterDot;}{g}---\left(11\right)$

Formula (11) is simplified, can be expressed as:

$T_{\max }=\frac{R_1}{R_1+R_2}{\mathrm{<figure-callout\; id="2"\; label="T"\; filenames="HDA0000129021630000022.png,HDA0000129021630000023.png"\; state="\{\{state\}\}">T</figure-callout>}}_2+\frac{R_2}{R_1+R_2}{\mathrm{<figure-callout\; id="1"\; label="T"\; filenames="HDA0000129021630000022.png,HDA0000129021630000032.png"\; state="\{\{state\}\}">T</figure-callout>}}_1+(R_m+\frac{R_1{R}_2}{R_1+R_2})G---\left(12\right)$

Wherein: G gives birth to heat, R for the inside of circle wall cylinder body ₁, R ₂, R _mFor justifying radially three thermal resistances of heat conduction steady temperature maximum value heat supply network network of wall cylinder body, R ₁Expression circle wall inner surface of cylinder block temperature T ₁With neutral temperature T _mBetween thermal resistance, R ₂Expression circle wall outer surface of cylinder block temperature T ₂With neutral temperature T _mBetween thermal resistance, R _mExpression circle wall cylinder body is heat conduction steady temperature maximum value T radially _MaxWith neutral temperature T _mBetween thermal resistance.R ₁, R ₂, R _mBe respectively:

${\mathrm{<figure-callout\; id="1"\; label="R"\; filenames="HDA0000129021630000022.png,HDA0000129021630000032.png"\; state="\{\{state\}\}">R</figure-callout>}}_1=\frac{1}{2\mathrm{\&pi;kL}}(\frac{1}{2}{f}_1-\frac{1}{2}-\ln r_1)---\left(13\right)$

${\mathrm{<figure-callout\; id="2"\; label="R"\; filenames="HDA0000129021630000022.png,HDA0000129021630000023.png"\; state="\{\{state\}\}">R</figure-callout>}}_2=\frac{1}{2\mathrm{\&pi;kL}}(-\frac{1}{2}{\mathrm{<figure-callout\; id="1"\; label="f"\; filenames="HDA0000129021630000022.png,HDA0000129021630000032.png"\; state="\{\{state\}\}">f</figure-callout>}}_1+\frac{1}{2}+\ln r_2)---\left(14\right)$

$R_m=\frac{[(\frac{1}{2}{f}_1-\frac{1}{2})({\mathrm{<figure-callout\; id="2"\; label="r"\; filenames="HDA0000129021630000022.png,HDA0000129021630000023.png"\; state="\{\{state\}\}">r</figure-callout>}}_2^2-{\mathrm{<figure-callout\; id="1"\; label="r"\; filenames="HDA0000129021630000022.png,HDA0000129021630000032.png"\; state="\{\{state\}\}">r</figure-callout>}}_1^2)+({\ln \mathrm{<figure-callout\; id="2"\; label="r"\; filenames="HDA0000129021630000022.png,HDA0000129021630000023.png"\; state="\{\{state\}\}">r</figure-callout>}}_2\&CenterDot;{\mathrm{<figure-callout\; id="1"\; label="r"\; filenames="HDA0000129021630000022.png,HDA0000129021630000032.png"\; state="\{\{state\}\}">r</figure-callout>}}_1^2-\mathrm{<figure-callout\; id="1"\; label="ln\; r"\; filenames="HDA0000129021630000022.png,HDA0000129021630000032.png"\; state="\{\{state\}\}">ln</figure-callout>}{r}_{}{}_1\&CenterDot;{\mathrm{<figure-callout\; id="2"\; label="r"\; filenames="HDA0000129021630000022.png,HDA0000129021630000023.png"\; state="\{\{state\}\}">r</figure-callout>}}_2^2)-(\frac{1}{2}{f}_1-\frac{1}{2}-\ln r_1)(-\frac{1}{2}{\mathrm{<figure-callout\; id="1"\; label="f"\; filenames="HDA0000129021630000022.png,HDA0000129021630000032.png"\; state="\{\{state\}\}">f</figure-callout>}}_1+\frac{1}{2}+\ln r_2)+{\mathrm{<figure-callout\; id="2"\; label="f"\; filenames="HDA0000129021630000022.png,HDA0000129021630000023.png"\; state="\{\{state\}\}">f</figure-callout>}}_2\ln \frac{r_2}{r_1}]}{4\mathrm{K\&pi;}({\mathrm{<figure-callout\; id="2"\; label="r"\; filenames="HDA0000129021630000022.png,HDA0000129021630000023.png"\; state="\{\{state\}\}">r</figure-callout>}}_2^2-{\mathrm{<figure-callout\; id="1"\; label="r"\; filenames="HDA0000129021630000022.png,HDA0000129021630000032.png"\; state="\{\{state\}\}">r</figure-callout>}}_1^2)L\&CenterDot;\ln \frac{r_2}{r_1}}---\left(15\right)$

$-\frac{(\frac{1}{2}{f}_1-\frac{1}{2}-\ln r_1)(-\frac{1}{2}{\mathrm{<figure-callout\; id="1"\; label="f"\; filenames="HDA0000129021630000022.png,HDA0000129021630000032.png"\; state="\{\{state\}\}">f</figure-callout>}}_1+\frac{1}{2}+{\ln r}_2)}{2\mathrm{K\&pi;L}\&CenterDot;\ln \frac{r_2}{{\mathrm{<figure-callout\; id="1"\; label="r"\; filenames="HDA0000129021630000022.png,HDA0000129021630000032.png"\; state="\{\{state\}\}">r</figure-callout>}}_1}}$

As shown in Figure 4 with above-mentioned corresponding radially heat conduction steady temperature maximum value heat supply network network model, three thermal resistance R ₁, R ₂And R _mBe the Y type and connect, just the first thermal resistance R ₁Right-hand member, the second thermal resistance R ₂Left end and the 3rd thermal resistance R _mLeft end all connect together, the temperature at the interface place of three thermal resistances is neutral temperature T _m, circle wall inner surface of cylinder block temperature T ₁At the first thermal resistance R ₁Left end, circle wall outer surface of cylinder block temperature T ₂At the second thermal resistance R ₂Right-hand member, temperature maximum T _MaxAt the 3rd thermal resistance R _mRight-hand member, the 3rd thermal resistance R _mRight-hand member connect the model that heat G is given birth in the inside of representative circle wall cylinder body.If research object is the structure that single round wall cylinder body is formed, its heat supply network network model is exactly the heat supply network network model of round wall cylinder body as shown in Figure 4; If the structure that research object is made up of a plurality of round wall cylinder bodies, then its heat supply network network is made up of several round wall cylinder body heat supply network networks connections shown in Figure 4.

According to the heat supply network network model of setting up, list corresponding thermal balance equation, as follows:

[Y][T]+[G]＝0 (16)

Wherein, [Y] is the thermal conductance matrix, and [T] is temperature node matrix, and the temperature node that is comprised is neutral temperature node and temperature maximum node, and [G] is the heat matrix.Through [T]=-[Y] ^-1[G], (finding the solution about the foundation of heat supply network network model thermal balance equation and steady temperature node can be with reference to P.H.Mellor, et al. can to calculate the steady temperature value of each temperature node in the heat supply network network model of being set up; " Lumped parameter thermal model for electrical machines of TEFC design; " Electric Power Applications, IEE Proceedings B, vol.138; Pp 205-218,1991).

In the heat supply network network model that research object is set up; Some can directly record the temperature on the inside and outside surface of circle wall cylinder body; Some need be through further calculating, specifically by the neutral temperature node in the temperature node matrix [T] that obtains and the steady temperature value of temperature maximum node, then according to the heat supply network network model that research object is set up; According to circuit in similar Kirchhoff's law, the temperature on the inside and outside surface of each circle wall cylinder body of confirming not record.

Step 2, according to each circle wall inner surface of cylinder block and hull-skin temperature T ₁, T ₂Judge radially heat conduction steady temperature maximum point r of each circle wall cylinder body _MaxDistribution.For the research object of forming by single round wall housing structure, its circle wall inner surface of cylinder block and hull-skin temperature T ₁, T ₂Be boundary condition.

According to circle wall inner surface of cylinder block and hull-skin temperature T ₁, T ₂Judge radially heat conduction steady temperature maximum point r _MaxThe concrete grammar of distribution situation is following:

The first step: equate as if circle wall inner surface of cylinder block and hull-skin temperature, i.e. T ₁=T ₂, temperature maximum point r then _MaxBe between round wall cylinder diameter and the external diameter, i.e. r _Max∈ (r ₁, r ₂).

The concrete foundation of judging is following:

Radially heat conduction steady temperature maximum point r with formula (5) gained _MaxThe expression formula two ends respectively divided by

With

Obtain following two equalities:

${\left(\frac{r_{\max }}{r_1}\right)}^2=\frac{1}{2}\&CenterDot;\frac{1}{\ln \frac{r_2}{{\mathrm{<figure-callout\; id="1"\; label="r"\; filenames="HDA0000129021630000022.png,HDA0000129021630000032.png"\; state="\{\{state\}\}">r</figure-callout>}}_1}}\&CenterDot;[{\left(\frac{r_2}{r_1}\right)}^2-1]---\left(17\right)$

${\left(\frac{r_{\max }}{r_2}\right)}^2=\frac{1}{2}\&CenterDot;\frac{1}{\ln \frac{r_1}{{\mathrm{<figure-callout\; id="2"\; label="r"\; filenames="HDA0000129021630000022.png,HDA0000129021630000023.png"\; state="\{\{state\}\}">r</figure-callout>}}_2}}\&CenterDot;[{\left(\frac{r_1}{r_2}\right)}^2-1]---\left(18\right)$

Through this conversion, can with in formula (17) and the formula (18) about r _MaxDistribution problem be converted into function

Field of definition x ∈ (0,1) ∪ (1 ,+∞) codomain problem.In Fig. 3, provide the codomain of this function under said field of definition of using MATLAB software to obtain.As can be seen from Figure 3; Function is discontinuous function; X=1 is its breakpoint; At this breakpoint place, the ultimate value of function is 1; When x ∈ (0,1), y＜1; When x ∈ (1,2), y＞1.To sum up, the T of working as is promptly arranged ₁=T ₂The time,

And

Can draw in this case temperature maximum point r thus _MaxBe between round wall cylinder diameter and the external diameter, i.e. r _Max∈ (r ₁, r ₂);

Second step: wall cylinder body surfaces externally and internally temperature is unequal as if justifying, i.e. T ₁≠ T ₂, then can pass through T ₁And T ₂Try to achieve temperature maximum point r _MaxDistribution decision content Δ and corresponding decision content upper limit Δ _MaxWith the lower limit Δ _Min, if

Temperature maximum T then _Max=max (T ₁, T ₂), the temperature maximum point is distributed in the inside surface or the outside surface of corresponding round wall cylinder body; If Δ ∈ (Δ _Min, Δ _Max) time, r then _Max∈ (r ₁, r ₂), temperature maximum point r just _MaxBe distributed in r ₁And r ₂Between.

Temperature maximum point r _MaxDistribution decision content Δ be shown below:

Δ＝|T ₂-T ₁| (19)

Further confirm decision content upper limit Δ _MaxWith decision content lower limit Δ _Min, detailed process is following:

By the radially heat conduction steady temperature maximum point r shown in the step 1 Chinese style (5) _MaxExpression formula can know that after structure, the material of circle wall cylinder body were confirmed, temperature maximum point was relevant with the difference of surfaces externally and internally temperature, then has:

${\mathrm{<figure-callout\; id="2"\; label="r"\; filenames="HDA0000129021630000022.png,HDA0000129021630000023.png"\; state="\{\{state\}\}">r</figure-callout>}}_2^2=\frac{2k}{\stackrel{\&CenterDot;}{g}}\&CenterDot;\frac{1}{\ln \frac{r_2}{r_1}}[{\mathrm{\&Delta;}}_{\max }+\frac{\stackrel{\&CenterDot;}{g}}{4k}({\mathrm{<figure-callout\; id="2"\; label="r"\; filenames="HDA0000129021630000022.png,HDA0000129021630000023.png"\; state="\{\{state\}\}">r</figure-callout>}}_2^2-{\mathrm{<figure-callout\; id="1"\; label="r"\; filenames="HDA0000129021630000022.png,HDA0000129021630000032.png"\; state="\{\{state\}\}">r</figure-callout>}}_1^2)]---\left(20\right)$

${\mathrm{<figure-callout\; id="1"\; label="r"\; filenames="HDA0000129021630000022.png,HDA0000129021630000032.png"\; state="\{\{state\}\}">r</figure-callout>}}_1^2=\frac{2k}{\stackrel{\&CenterDot;}{g}}\&CenterDot;\frac{1}{\ln \frac{r_2}{r_1}}[{\mathrm{\&Delta;}}_{\min }+\frac{\stackrel{\&CenterDot;}{g}}{4k}({\mathrm{<figure-callout\; id="2"\; label="r"\; filenames="HDA0000129021630000022.png,HDA0000129021630000023.png"\; state="\{\{state\}\}">r</figure-callout>}}_2^2-{\mathrm{<figure-callout\; id="1"\; label="r"\; filenames="HDA0000129021630000022.png,HDA0000129021630000032.png"\; state="\{\{state\}\}">r</figure-callout>}}_1^2)]---\left(21\right)$

Can obtain decision content upper limit Δ by last two formulas _MaxWith decision content lower limit Δ _MinFor:

$\left\{\begin{array}{c}{\mathrm{\&Delta;}}_{\max }=|\frac{\stackrel{\&CenterDot;}{g}}{2k}\&CenterDot;\ln \frac{r_2}{{\mathrm{<figure-callout\; id="1"\; label="r"\; filenames="HDA0000129021630000022.png,HDA0000129021630000032.png"\; state="\{\{state\}\}">r</figure-callout>}}_1}\&CenterDot;{\mathrm{<figure-callout\; id="2"\; label="r"\; filenames="HDA0000129021630000022.png,HDA0000129021630000023.png"\; state="\{\{state\}\}">r</figure-callout>}}_2^2-\frac{\stackrel{\&CenterDot;}{g}}{4k}({\mathrm{<figure-callout\; id="2"\; label="r"\; filenames="HDA0000129021630000022.png,HDA0000129021630000023.png"\; state="\{\{state\}\}">r</figure-callout>}}_2^2-{\mathrm{<figure-callout\; id="1"\; label="r"\; filenames="HDA0000129021630000022.png,HDA0000129021630000032.png"\; state="\{\{state\}\}">r</figure-callout>}}_1^2)|\\ {\mathrm{\&Delta;}}_{\min }=|\frac{\stackrel{\&CenterDot;}{g}}{2k}\&CenterDot;\ln \frac{r_2}{{\mathrm{<figure-callout\; id="1"\; label="r"\; filenames="HDA0000129021630000022.png,HDA0000129021630000032.png"\; state="\{\{state\}\}">r</figure-callout>}}_1}\&CenterDot;{\mathrm{<figure-callout\; id="1"\; label="r"\; filenames="HDA0000129021630000022.png,HDA0000129021630000032.png"\; state="\{\{state\}\}">r</figure-callout>}}_1^2-\frac{\stackrel{\&CenterDot;}{g}}{4k}({\mathrm{<figure-callout\; id="2"\; label="r"\; filenames="HDA0000129021630000022.png,HDA0000129021630000023.png"\; state="\{\{state\}\}">r</figure-callout>}}_2^2-{\mathrm{<figure-callout\; id="1"\; label="r"\; filenames="HDA0000129021630000022.png,HDA0000129021630000032.png"\; state="\{\{state\}\}">r</figure-callout>}}_1^2)|\end{array}\right.---\left(22\right)$

Step 3, according to the distribution situation of the radially heat conduction steady temperature maximum point in the step 2, determining whether needs approximate compensation tache Δ T, confirms heat conduction steady temperature maximum value.

Because item has carried out the first-order linear match to

in the step 1, the effect of compensation tache Δ T is exactly the error that compensation brings thus.Radially the process of specifically setting up of the compensation tache Δ T of heat conduction steady temperature maximum value heat supply network network model is following:

The first step: the distribution according to the radially heat conduction steady temperature maximum point of obtaining in the step 2 judges whether that needs carry out temperature compensation:

When heat conduction steady temperature maximum point radially is distributed in round wall cylinder body surperficial, the radially heat conduction steady temperature maximum value T of circle wall cylinder body _Max=max (T ₁, T ₂), need not carry out temperature compensation this moment;

When radially heat conduction steady temperature maximum point is distributed between round wall cylinder diameter and the external diameter, need carry out temperature compensation, the method for compensation is following:

Can know that by step 1 the error e that is produced by the first-order linear match can be expressed from the next:

$e=\frac{1}{2}{c}_1\ln \left(\frac{2k}{\stackrel{\&CenterDot;}{g}}{c}_1\right)-(\frac{1}{2}{f}_1{c}_1+\frac{1}{2}\frac{\stackrel{\&CenterDot;}{g}}{2k}{f}_2)---\left(23\right)$

Wherein: $c_1=\frac{1}{\mathrm{Ln}\frac{r_2}{r_1}}[(T_2-T_1)+\frac{\stackrel{\&CenterDot;}{g}}{4k}({\mathrm{<figure-callout\; id="2"\; label="r"\; filenames="HDA0000129021630000022.png,HDA0000129021630000023.png"\; state="\{\{state\}\}">r</figure-callout>}}_2^2-{\mathrm{<figure-callout\; id="1"\; label="r"\; filenames="HDA0000129021630000022.png,HDA0000129021630000032.png"\; state="\{\{state\}\}">r</figure-callout>}}_1^2)].$

Make compensation tache Δ T=e, then final round wall cylinder body is heat conduction steady temperature maximum value T ' radially _MaxFor:

T′ _max＝T _max+ΔT (24)

Wherein, T _MaxObtain by step 1.

What then obtain according to formula (24) is exactly radially heat conduction steady temperature maximum value of final round wall cylinder body to be determined.

Add compensation tache Δ T to step 1 and set up in the heat supply network network model, it is as shown in Figure 5 to obtain complete radially heat conduction steady temperature maximum value heat supply network network model, in heat supply network network model shown in Figure 4, adds compensation tache Δ T.

Embodiment

With stator core in the electric machine structure and stator teeth is object; Use radially heat conduction steady temperature maximum value heat supply network network modeling method provided by the present invention; And use the mellor modeling method to set up its heat supply network network model respectively; And the solving result of each self model policy result with ANSYS software compared, checking uses modeling method provided by the invention to set up the correctness of model, and then proves the practicality and the validity of modeling method provided by the invention.Motor stator core as embodiment is as shown in Figure 6 with the stator teeth structure, and wherein the value of each parameter is as shown in table 1.

Table 1 is used for the motor stator core and the tooth portion parameter of modeling

Parameter	Value	Unit
			Stator core internal diameter r 1_iron	0.0828	m
Stator core external diameter r 2_iron	0.1016	m
			The stator core length L iron	0.1207	m
Stator core heat-conduction coefficient k iron	45.2	w/(m·℃)
			The inner living hot G of stator core iron	141.9579	w
Stator teeth internal diameter r 1_teeth	0.0635	m
			Stator teeth external diameter r 2_teeth	0.0828	m
The stator teeth length L teeth	0.1207	m
			Stator teeth heat-conduction coefficient k teeth	45.2	w/(m·℃)
The inner living hot G of stator teeth teeth	141.9622	w
			Stator teeth tooth pitch φ p	10	o
Stator teeth groove width φ e	5.6	o

For present embodiment, can know the temperature outside T of stator core according to Fig. 6 _{2_iorn}And the inboard temperature T of stator teeth _{1_teeth}All belong to hull-skin temperature, can be through measuring, and below can be used as to set up the boundary condition of heat supply network network model.In the present embodiment, the measured boundary condition T that obtains _{2_iorn}=20.4 ℃, T _{1_teeth}=20 ℃.

1, use modeling method provided by the present invention to confirm the radially steady temperature maximal value of embodiment:

The first step: use heat conduction steady temperature maximum value heat supply network network radially to set up the heat supply network network model of embodiment:

At first set up the heat supply network network model framework of embodiment, as shown in Figure 7.This model is made up of the heat supply network network of two round wall housing structures, respectively corresponding stator core and stator teeth; Can know that through Fig. 6 the inner surface of embodiment modeling object stator core is close together with the outer surface of stator teeth, so think that in modeling these two surface temperatures are identical, both T among Fig. 7 _{2_teeth}(T _{1_iron}).

After having set up the heat supply network network model framework of embodiment; Stator core and each self-corresponding thermal resistance of stator teeth in the embodiment heat supply network network model calculated in parameter substitution formula (13)～(15) of table 1, it should be noted that after the corresponding thermal resistance calculation of stator teeth is come out and multiply by coefficient (φ _p/ φ _e); This is because stator teeth is not a complete round wall housing structure, on its circumference, is being evenly distributed in order to settle the stator slot of stator winding, so its volume is than the round wall cylinder body of same radius and length; Reduce the stator slot part, therefore will multiply by this coefficient.Result after the calculating is following:

$\left\{\begin{array}{c}{\mathrm{<figure-callout\; id="1"\; label="R"\; filenames="HDA0000129021630000022.png,HDA0000129021630000032.png"\; state="\{\{state\}\}">R</figure-callout>}}_{1\_\mathrm{iron}}=0.00322\\ {\mathrm{<figure-callout\; id="2"\; label="R"\; filenames="HDA0000129021630000022.png,HDA0000129021630000023.png"\; state="\{\{state\}\}">R</figure-callout>}}_{2\_\mathrm{iron}}=0.00274\\ R_{m\_\mathrm{iron}}=-0.00049\end{array}\right.$

$\left\{\begin{array}{c}{\mathrm{<figure-callout\; id="1"\; label="R"\; filenames="HDA0000129021630000022.png,HDA0000129021630000032.png"\; state="\{\{state\}\}">R</figure-callout>}}_{1\_\mathrm{teeth}}=0.00763\\ {\mathrm{<figure-callout\; id="2"\; label="R"\; filenames="HDA0000129021630000022.png,HDA0000129021630000023.png"\; state="\{\{state\}\}">R</figure-callout>}}_{2\_\mathrm{teeth}}=0.00619\\ R_{m\_\mathrm{teeth}}=-0.00113\end{array}\right.$

List its thermal balance equation according to the heat supply network network model of embodiment among Fig. 7 then, as follows:

[Y][T]＝-[G]

Wherein:

$\left[Y\right]=\left[\begin{array}{cccc}\frac{-1}{R_{m\_\mathrm{teeth}}}-\frac{1}{({\mathrm{<figure-callout\; id="2"\; label="R"\; filenames="HDA0000129021630000022.png,HDA0000129021630000023.png"\; state="\{\{state\}\}">R</figure-callout>}}_{2\_\mathrm{teeth}}+{\mathrm{<figure-callout\; id="1"\; label="R"\; filenames="HDA0000129021630000022.png,HDA0000129021630000032.png"\; state="\{\{state\}\}">R</figure-callout>}}_{1\_\mathrm{iron}})}-\frac{1}{{\mathrm{<figure-callout\; id="1"\; label="R"\; filenames="HDA0000129021630000022.png,HDA0000129021630000032.png"\; state="\{\{state\}\}">R</figure-callout>}}_{1\_\mathrm{teeth}}}& \frac{1}{R_{m\_\mathrm{teeth}}}& \frac{1}{({\mathrm{<figure-callout\; id="2"\; label="R"\; filenames="HDA0000129021630000022.png,HDA0000129021630000023.png"\; state="\{\{state\}\}">R</figure-callout>}}_{2\_\mathrm{teeth}}+{\mathrm{<figure-callout\; id="1"\; label="R"\; filenames="HDA0000129021630000022.png,HDA0000129021630000032.png"\; state="\{\{state\}\}">R</figure-callout>}}_{1\_\mathrm{iron}})}& 0\\ \frac{1}{R_{m\_\mathrm{teeth}}}& \frac{-1}{R_{m\_\mathrm{<figure-callout\; id="0"\; label="teeth"\; filenames="HDA0000129021630000022.png"\; state="\{\{state\}\}">teeth</figure-callout>}}}& 0& 0\\ \frac{1}{({\mathrm{<figure-callout\; id="2"\; label="R"\; filenames="HDA0000129021630000022.png,HDA0000129021630000023.png"\; state="\{\{state\}\}">R</figure-callout>}}_{2\_\mathrm{teeth}}+{\mathrm{<figure-callout\; id="1"\; label="R"\; filenames="HDA0000129021630000022.png,HDA0000129021630000032.png"\; state="\{\{state\}\}">R</figure-callout>}}_{1\_\mathrm{iron}})}& 0& \frac{-1}{({\mathrm{<figure-callout\; id="2"\; label="R"\; filenames="HDA0000129021630000022.png,HDA0000129021630000023.png"\; state="\{\{state\}\}">R</figure-callout>}}_{2\_\mathrm{teeth}}+{\mathrm{<figure-callout\; id="1"\; label="R"\; filenames="HDA0000129021630000022.png,HDA0000129021630000032.png"\; state="\{\{state\}\}">R</figure-callout>}}_{1\_\mathrm{iron}})}-\frac{1}{R_{m\_\mathrm{iron}}}-\frac{1}{{\mathrm{<figure-callout\; id="2"\; label="R"\; filenames="HDA0000129021630000022.png,HDA0000129021630000023.png"\; state="\{\{state\}\}">R</figure-callout>}}_{2\_\mathrm{iron}}}& \frac{1}{R_{m\_\mathrm{<figure-callout\; id="0"\; label="iron"\; filenames="HDA0000129021630000022.png"\; state="\{\{state\}\}">iron</figure-callout>}}}\\ 0& 0& \frac{1}{R_{m\_\mathrm{iron}}}& \frac{-1}{R_{m\_\mathrm{iron}}}\end{array}\right],$

$\left[T\right]=\left[\begin{array}{c}{T}_{m\_\mathrm{teeth}}\\ T_{\max \_\mathrm{teeth}}\\ T_{m\_\mathrm{iron}}\\ T_{\max \_\mathrm{iron}}\end{array}\right],$ $\left[G\right]=\left[\begin{array}{c}{\mathrm{<figure-callout\; id="1"\; label="T"\; filenames="HDA0000129021630000022.png,HDA0000129021630000032.png"\; state="\{\{state\}\}">T</figure-callout>}}_{1\_\mathrm{teeth}}/{\mathrm{<figure-callout\; id="1"\; label="R"\; filenames="HDA0000129021630000022.png,HDA0000129021630000032.png"\; state="\{\{state\}\}">R</figure-callout>}}_{1\_\mathrm{teeth}}\\ {\mathrm{<figure-callout\; id="2"\; label="G\; teeth\; T"\; filenames="HDA0000129021630000022.png,HDA0000129021630000023.png"\; state="\{\{state\}\}">G</figure-callout>}}_{\mathrm{teeth}}& \\ T_{}{}_{2\_\mathrm{iron}}/{\mathrm{<figure-callout\; id="2"\; label="R"\; filenames="HDA0000129021630000022.png,HDA0000129021630000023.png"\; state="\{\{state\}\}">R</figure-callout>}}_{2\_\mathrm{iron}}\\ G_{\mathrm{iron}}\end{array}\right].$

Find the solution thermal balance equation, obtain the temperature matrix $\left[T\right]=\left[\begin{array}{c}{T}_{m\_\mathrm{Teeth}}\\ T_{\mathrm{Max}\_\mathrm{Teeth}}\\ T_{m\_\mathrm{Iron}}\\ T_{\mathrm{Max}\_\mathrm{Iron}}\end{array}\right]=\left[\begin{array}{c}20.97029\\ 20.80945\\ 20.83071\\ 20.76075\end{array}\right].$

Second step: judge r _{Max_teeth}And r _{Max_iron}Distribution:

The stator core of obtaining through the first step and the neutral temperature T of stator teeth _{M_teeth}, T _{M_iron}Calculate stator core inboard (the stator teeth outside) temperature T _{2_teeth}(T _{1_iron}).According to circuit in similar Kirchhoff's law, can get:

T _{2_teeth}(T _{1_iron})＝R _{1_iron}/(R _{1_iron}+R _{2_teeth})*T _{m_teeth}+R _{2_teeth}/(R1 _{_iron}+R _{2_teeth})*T _{m_iron}＝20.87847

Calculate stator core and stator teeth temperature maximum of points r _{Max_iron}And r _{Max_teeth}The maximal value of judgment threshold and minimum value:

$\left\{\begin{array}{c}{\mathrm{\&Delta;}}_{\max \_\mathrm{teeth}}=1.06772\\ {\mathrm{\&Delta;}}_{\min \_\mathrm{teeth}}=0.89491\end{array}\right.$

$\left\{\begin{array}{c}{\mathrm{\&Delta;}}_{\max \_\mathrm{iron}}=0.45250\\ {\mathrm{\&Delta;}}_{\min \_\mathrm{iron}}=0.39486\end{array}\right.$

Calculate the radially judgment threshold Δ of steady temperature maximum of points of stator core _Iron=| T _{2_iron}-T _{1_iron}|=0.47847, because Δ _Iron＞Δ _{Max_iron}So, its radially the steady temperature maximum of points be distributed in the surface of stator core, again because of T _{1_iron}＞T _{2_iron}So, r _{Max_iron}=r _{1_iron}, and then the radially steady temperature maximum of T of stator core ' _{Max_iron}=T _{1_iron}=20.8785.

In like manner calculate the radially judgment threshold Δ of steady temperature maximum of points of stator teeth _Teeth=| T _{2_teeth}-T _{1_teeth}|=0.8785, because Δ _{Max_teeth}＞Δ _Teeth＞Δ _{Min_teeth}So, its radially the steady temperature maximum of points be distributed in (r _{1_teeth}, r _{2_teeth}) between.

The 3rd step: calculate compensation tache.

Because stator core radially steady temperature maximal value is distributed in its surface, thus its radially the steady temperature maximal value just equal its surface temperature, not be used in and compensate, a compensation tache that therefore only needs to calculate stator teeth gets final product.With boundary condition T _{1_teeth}And the hull-skin temperature T of tooth portion that goes out through heat supply network network Model Calculation _{2_teeth}Be updated in the computing formula (23) of compensation tache, can get:

$\mathrm{\&Delta;T}=e=\frac{1}{2}{c}_1\ln \left(\frac{2k}{\stackrel{\&CenterDot;}{g}}{c}_1\right)-(\frac{1}{2}{f}_1{c}_1+\frac{1}{2}\frac{\stackrel{\&CenterDot;}{g}}{2k}{f}_2)=0.0762$

Then stator teeth radially the steady temperature maximum of T ' _{Max_teeth}=T _{Max_teeth}+ Δ T=20.8857.

Can get thus stator core and tooth portion radially the steady temperature maximal value be respectively:

T′ _{max_teeth}＝20.8785℃

T′ _{max_teeth}＝20.8857℃

2, use the mellor modeling method to confirm the radially steady temperature value of embodiment:

The embodiment heat supply network network model structure of using the mellor method to build is identical with the heat supply network network model structure of using method provided by the present invention to build, and the difference of two heat supply network network models is that the corresponding thermal resistance calculation method of stator core and stator teeth is different.

The computing method of the corresponding thermal resistance of circle wall housing structure are following in the mellor method:

$\left\{\begin{array}{c}{\mathrm{<figure-callout\; id="1"\; label="R"\; filenames="HDA0000129021630000022.png,HDA0000129021630000032.png"\; state="\{\{state\}\}">R</figure-callout>}}_1=\frac{1}{4\mathrm{\&pi;kL}}[\frac{{2\mathrm{<figure-callout\; id="2"\; label="r"\; filenames="HDA0000129021630000022.png,HDA0000129021630000023.png"\; state="\{\{state\}\}">r</figure-callout>}}_2^2\ln ({\mathrm{<figure-callout\; id="2"\; label="r"\; filenames="HDA0000129021630000022.png,HDA0000129021630000023.png"\; state="\{\{state\}\}">r</figure-callout>}}_2/r_1)}{{\mathrm{<figure-callout\; id="2"\; label="r"\; filenames="HDA0000129021630000022.png,HDA0000129021630000023.png"\; state="\{\{state\}\}">r</figure-callout>}}_2^2-{\mathrm{<figure-callout\; id="1"\; label="r"\; filenames="HDA0000129021630000022.png,HDA0000129021630000032.png"\; state="\{\{state\}\}">r</figure-callout>}}_1^2}-1]\\ {\mathrm{<figure-callout\; id="2"\; label="R"\; filenames="HDA0000129021630000022.png,HDA0000129021630000023.png"\; state="\{\{state\}\}">R</figure-callout>}}_2=\frac{1}{4\mathrm{\&pi;kL}}[1-\frac{{2\mathrm{<figure-callout\; id="1"\; label="r"\; filenames="HDA0000129021630000022.png,HDA0000129021630000032.png"\; state="\{\{state\}\}">r</figure-callout>}}_1^2\ln ({\mathrm{<figure-callout\; id="2"\; label="r"\; filenames="HDA0000129021630000022.png,HDA0000129021630000023.png"\; state="\{\{state\}\}">r</figure-callout>}}_2/r_1)}{{\mathrm{<figure-callout\; id="2"\; label="r"\; filenames="HDA0000129021630000022.png,HDA0000129021630000023.png"\; state="\{\{state\}\}">r</figure-callout>}}_2^2-{\mathrm{<figure-callout\; id="1"\; label="r"\; filenames="HDA0000129021630000022.png,HDA0000129021630000032.png"\; state="\{\{state\}\}">r</figure-callout>}}_1^2}]\\ R_m=\frac{-1}{8\mathrm{\&pi;kL}({\mathrm{<figure-callout\; id="2"\; label="r"\; filenames="HDA0000129021630000022.png,HDA0000129021630000023.png"\; state="\{\{state\}\}">r</figure-callout>}}_2^2-{\mathrm{<figure-callout\; id="1"\; label="r"\; filenames="HDA0000129021630000022.png,HDA0000129021630000032.png"\; state="\{\{state\}\}">r</figure-callout>}}_1^2)}[{\mathrm{<figure-callout\; id="1"\; label="r"\; filenames="HDA0000129021630000022.png,HDA0000129021630000032.png"\; state="\{\{state\}\}">r</figure-callout>}}_1^2+{\mathrm{<figure-callout\; id="2"\; label="r"\; filenames="HDA0000129021630000022.png,HDA0000129021630000023.png"\; state="\{\{state\}\}">r</figure-callout>}}_2^2-\frac{{4\mathrm{<figure-callout\; id="1"\; label="r"\; filenames="HDA0000129021630000022.png,HDA0000129021630000032.png"\; state="\{\{state\}\}">r</figure-callout>}}_1^2{\mathrm{<figure-callout\; id="2"\; label="r"\; filenames="HDA0000129021630000022.png,HDA0000129021630000023.png"\; state="\{\{state\}\}">r</figure-callout>}}_2^2\ln ({\mathrm{<figure-callout\; id="2"\; label="r"\; filenames="HDA0000129021630000022.png,HDA0000129021630000023.png"\; state="\{\{state\}\}">r</figure-callout>}}_2/r_1)}{{\mathrm{<figure-callout\; id="2"\; label="r"\; filenames="HDA0000129021630000022.png,HDA0000129021630000023.png"\; state="\{\{state\}\}">r</figure-callout>}}_2^2-{\mathrm{<figure-callout\; id="1"\; label="r"\; filenames="HDA0000129021630000022.png,HDA0000129021630000032.png"\; state="\{\{state\}\}">r</figure-callout>}}_1^2}]\end{array}\right.$

What the mellor method was calculated is the medial temperature of circle wall housing structure, and computing method are following:

$T_{\mathrm{mean}}=\frac{R_1}{{\mathrm{<figure-callout\; id="1"\; label="R"\; filenames="HDA0000129021630000022.png,HDA0000129021630000032.png"\; state="\{\{state\}\}">R</figure-callout>}}_1+R_2}{\mathrm{<figure-callout\; id="2"\; label="T"\; filenames="HDA0000129021630000022.png,HDA0000129021630000023.png"\; state="\{\{state\}\}">T</figure-callout>}}_2+\frac{R_2}{{\mathrm{<figure-callout\; id="1"\; label="R"\; filenames="HDA0000129021630000022.png,HDA0000129021630000032.png"\; state="\{\{state\}\}">R</figure-callout>}}_1+R_2}{\mathrm{<figure-callout\; id="1"\; label="T"\; filenames="HDA0000129021630000022.png,HDA0000129021630000032.png"\; state="\{\{state\}\}">T</figure-callout>}}_1+(R_m+\frac{R_1{R}_2}{{\mathrm{<figure-callout\; id="1"\; label="R"\; filenames="HDA0000129021630000022.png,HDA0000129021630000032.png"\; state="\{\{state\}\}">R</figure-callout>}}_1+R_2})G$

Following according to stator core in the above-mentioned mellor method calculating present embodiment with the corresponding thermal resistance of stator teeth:

$\left\{\begin{array}{c}{\mathrm{<figure-callout\; id="1"\; label="R"\; filenames="HDA0000129021630000022.png,HDA0000129021630000032.png"\; state="\{\{state\}\}">R</figure-callout>}}_{1\_\mathrm{iron}}=0.00278\\ {\mathrm{<figure-callout\; id="2"\; label="R"\; filenames="HDA0000129021630000022.png,HDA0000129021630000023.png"\; state="\{\{state\}\}">R</figure-callout>}}_{2\_\mathrm{iron}}=0.00319\\ R_{m\_\mathrm{iron}}=-0.00947\end{array}\right.$

$\left\{\begin{array}{c}{\mathrm{<figure-callout\; id="1"\; label="R"\; filenames="HDA0000129021630000022.png,HDA0000129021630000032.png"\; state="\{\{state\}\}">R</figure-callout>}}_{1\_\mathrm{teeth}}=0.00752\\ {\mathrm{<figure-callout\; id="2"\; label="R"\; filenames="HDA0000129021630000022.png,HDA0000129021630000023.png"\; state="\{\{state\}\}">R</figure-callout>}}_{2\_\mathrm{teeth}}=0.00630\\ R_{m\_\mathrm{teeth}}=-0.00446\end{array}\right.$

Stator core that then calculates and stator teeth medial temperature are respectively:

T _{mean_iron}＝19.5431℃

T _{mean_teeth}＝20.3527℃

3, ANSYS emulation contrast:

Fig. 8 is a radially steady temperature distribution curve of this instance stator core that emulation obtains in ANSYS, and Fig. 9 is a radially steady temperature distribution curve of stator teeth, and horizontal ordinate is a radius among two width of cloth figure, and ordinate is the pairing temperature of radius.Can know that from Fig. 8, Fig. 9 the radially steady temperature maximal value of stator core is distributed in the inboard of iron core, value is 20.911 ℃; The radially steady temperature maximal value of stator teeth is distributed in (r _{1_teeth}, r _{2_teeth}) between, value is 20.916 ℃.

Using stator core maximum temperature that the heat supply network network modeling method of radially heat conduction steady temperature maximum value provided by the invention tries to achieve is 0.16% with the maximum temperature error of ANSYS simulation result, and the stator teeth maximum temperature is 0.14% with the maximum temperature error of ANSYS simulation result; Using stator core temperature that mellor tries to achieve is 6.54% with the maximum temperature error of ANSYS simulation result, and the stator teeth temperature is 2.69% with the maximum temperature error of ANSYS simulation result.This shows; Use the present invention to provide the definite radially steady temperature maximal value of method than using the definite temperature of mellor method more near ANSYS emulation gained result; And with respect to the ANSYS software emulation; Use method provided by the present invention, confirm that radially steady temperature maximum value required time is shorter.

Claims (4)

1. confirm the radially heat supply network network modeling method of heat conduction steady temperature maximum value for one kind, it is characterized in that, may further comprise the steps:

Step 1, at first use heat conduction steady temperature maximum value network radially to set up the heat supply network network model of institute's research object, specifically: the structure to each circle wall cylinder body is formed is provided with three thermal resistance R ₁, R ₂And R _m, the first thermal resistance R ₁Right-hand member, the second thermal resistance R ₂Left end and the 3rd thermal resistance R _mLeft end all connect together, three thermal resistances are the Y type and connect, the temperature at interface place is neutral temperature T _m, circle wall inner surface of cylinder block temperature T ₁At the first thermal resistance R ₁Left end, circle wall outer surface of cylinder block temperature T ₂At the second thermal resistance R ₂Right-hand member, temperature maximum T _MaxAt the 3rd thermal resistance R _mRight-hand member, the 3rd thermal resistance R _mThe right-hand member inside that connects circle wall cylinder body give birth to heat G; If the structure that research object is made up of plural round wall cylinder body, then its heat supply network network model is made up of the heat supply network network model connection of each circle wall cylinder body;

According to the heat supply network network model of the research object of being set up, list thermal balance equation then: [Y] [T]+[G]=0, wherein, [Y] is the thermal conductance matrix, [T] is temperature node matrix, [G] is the heat matrix, through [T]=-[Y] ^-1[G], the neutral temperature T in the heat supply network network model that obtains being set up _mWith temperature maximum T _Max,, confirm the inside and outside surface temperature of each circle wall cylinder body further according to the heat supply network network model of the research object of being set up;

Step 2, confirm the radially heat conduction steady temperature maximum point r of each circle wall cylinder body _MaxDistribution: (1) is if circle wall inner surface of cylinder block temperature T ₁With circle wall outer surface of cylinder block temperature T ₂Equate, then temperature maximum point r _MaxBe between round wall cylinder diameter and the external diameter; (2) if circle wall inner surface of cylinder block temperature T ₁With circle wall outer surface of cylinder block temperature T ₂Unequal, according to temperature maximum point r _MaxDistribution decision content Δ=| T ₂-T ₁| and corresponding decision content upper limit Δ _MaxWith the lower limit Δ _Min, confirm temperature maximum point r _MaxDistribution: if

The temperature maximum point is distributed in the inside surface or the outside surface of round wall cylinder body; If Δ ∈ (Δ _Min, Δ _Max) time, temperature maximum point r then _MaxBe distributed in r ₁And r ₂Between: r _Max∈ (r ₁, r ₂);

Step 3, if heat conduction steady temperature maximum point r radially _MaxBe distributed on the inside surface or outside surface of round wall cylinder body then final heat conduction steady temperature maximum value T ' _Max=max (T ₁, T ₂); If heat conduction steady temperature maximum point r radially _MaxBe distributed between the internal diameter and external diameter of round wall cylinder body, then confirm approximate compensation tache Δ T, confirm final heat conduction steady temperature maximum value T ' then _Max=T _Max+ Δ T.

2. a kind of radially heat supply network network modeling method of heat conduction steady temperature maximum value of confirming according to claim 1 is characterized in that, the temperature maximum T in the heat supply network network model of each the circle wall cylinder body described in the step 1 _Max, three thermal resistance R ₁, R ₂And R _mExpression formula obtain through following process:

(1) radially heat conduction steady-state heat balance equation is:

$\frac{\&PartialD;T^2\left(r\right)}{\&PartialD;r^2}+\frac{1}{r}\frac{\&PartialD;T\left(r\right)}{\&PartialD;r}+\frac{\stackrel{\&CenterDot;}{g}}{k}=0---\left(1\right)$

Wherein, T (r) is radially heat conduction Temperature Distribution of circle wall cylinder body function; R representes the radius radially of circle wall cylinder body;

is the heat-conduction coefficient of circle wall cylinder block material for the heat generation rate of circle wall cylinder body unit volume, k;

The general solution of the radially heat conduction steady-state heat balance equation of formula (1) is:

$T\left(r\right)=-\frac{\stackrel{\&CenterDot;}{g}}{4k}{r}^2+c_1\ln r+c_2---\left(2\right)$

Wherein, c ₁And c ₂Be constant,, make that derivative is 0, obtain stationary point r formula (2) differentiate ₀: With stationary point r ₀The second derivative of substitution formula (2) obtains

Because second derivative is located less than 0, so stationary point r in the stationary point ₀Be maximum point r: $r_{\mathrm{Max}}^2=r_0^2=\frac{2k}{\stackrel{\&CenterDot;}{g}}\&CenterDot;c_1;$

(2) establish boundary condition T (r ₁)=T ₁, T (r ₂)=T ₂, T ₁, T ₂Be respectively round wall inner surface of cylinder block temperature and hull-skin temperature, r ₁, r ₂Be respectively the internal diameter and the external diameter of round wall cylinder body, boundary condition is brought in the formula (2), obtain constant c ₁And c ₂Expression formula following:

$c_1=\frac{1}{\ln \frac{r_2}{r_1}}[(T_2-T_1)+\frac{\stackrel{\&CenterDot;}{g}}{4k}(r_2^2-r_1^2)]---\left(3\right)$

$c_2=\frac{1}{\ln \frac{r_2}{r_1}}(\ln r_2\&CenterDot;T_1-\ln r_1\&CenterDot;T_2)+\frac{\stackrel{\&CenterDot;}{g}}{4k}\&CenterDot;\frac{1}{\ln \frac{r_2}{r_1}}({\ln r}_2\&CenterDot;r_1^2-\ln r_1\&CenterDot;r_2^2)---\left(4\right)$

With c ₁Expression formula bring into

Obtain radially heat conduction steady temperature maximum point r _Max:

$r_{\max }^2=\frac{2k}{\stackrel{\&CenterDot;}{g}}\&CenterDot;\frac{1}{\ln \frac{r_2}{r_1}}[(T_2-T_1)+\frac{\stackrel{\&CenterDot;}{g}}{4k}(r_2^2-r_1^2)]---\left(5\right)$

(3) with formula (5) substitution formula (2), obtain temperature maximum T _Max:

$T_{\max }=-\frac{1}{2}{c}_1+c_1\ln {(\frac{2k}{\stackrel{\&CenterDot;}{g}}\&CenterDot;c_1)}^{\frac{1}{2}}+c_2---\left(6\right)$

(4)

in the formula (6) done linear transformation, it is transformed to:

$c_1\ln {(\frac{2k}{\stackrel{\&CenterDot;}{g}}\&CenterDot;c_1)}^{\frac{1}{2}}=\frac{\stackrel{\&CenterDot;}{g}}{4k}\&CenterDot;(\frac{2k}{\stackrel{\&CenterDot;}{g}}\&CenterDot;c_1)\&CenterDot;\ln (\frac{2k}{\stackrel{\&CenterDot;}{g}}\&CenterDot;c_1)---\left(7\right)$

Make with in the x substitution formula (7), get following formula:

$c_1\ln {(\frac{2k}{\stackrel{\&CenterDot;}{g}}\&CenterDot;c_1)}^{\frac{1}{2}}=\frac{\stackrel{\&CenterDot;}{g}}{4k}\&CenterDot;x\&CenterDot;\ln x,x\&Element;[r_1^2,r_2^2]---\left(8\right)$

Function xlnx is carried out the first-order linear match in

scope, the result after the match is:

x·lnx≈f ₁x+f ₂ (9)

f ₁And f ₂The coefficient that obtains for the first-order linear match;

In the linear transform with the substitution as a result after the xlnx match , can get:

$c_1\ln {(\frac{2k}{\stackrel{\&CenterDot;}{g}}\&CenterDot;c_1)}^{\frac{1}{2}}\&ap;\frac{1}{2}{f}_1\&CenterDot;c_1+\frac{\stackrel{\&CenterDot;}{g}}{4k}\&CenterDot;f_2---\left(10\right)$

(5) with formula (10), formula (3) and formula (4) substitution formula (6), obtain temperature maximum T _Max:

$T_{\max }=\frac{(\frac{1}{2}{f}_1-\frac{1}{2}-\ln r_1)}{\ln \frac{r_2}{r_1}}{T}_2+\frac{(-\frac{1}{2}{f}_1+\frac{1}{2}+\ln r_2)}{\ln \frac{r_2}{r_1}}{T}_1+[\frac{(\frac{1}{2}{f}_1-\frac{1}{2})(r_2^2-r_1^2)+({\ln r}_2\&CenterDot;r_1^2-\ln r_1\&CenterDot;r_2^2)}{\ln \frac{r_2}{r_1}\&CenterDot;4k}+\frac{f_2}{4k}]\&CenterDot;\stackrel{\&CenterDot;}{g}---\left(11\right)$

Formula (11) is simplified, is obtained:

$T_{\max }=\frac{R_1}{R_1+R_2}{T}_2+\frac{R_2}{R_1+R_2}{T}_1+(R_m+\frac{R_1{R}_2}{R_1+R_2})G---\left(12\right)$

Three thermal resistance R ₁, R ₂, R _mBe respectively:

$R_1=\frac{1}{2\mathrm{\&pi;kL}}(\frac{1}{2}{f}_1-\frac{1}{2}-\ln r_1)---\left(13\right)$

$R_2=\frac{1}{2\mathrm{\&pi;kL}}(-\frac{1}{2}{f}_1+\frac{1}{2}+\ln r_2)---\left(14\right)$

$R_m=\frac{[(\frac{1}{2}{f}_1-\frac{1}{2})(r_2^2-r_1^2)+({\ln r}_2\&CenterDot;r_1^2-\ln r_1\&CenterDot;r_2^2)-(\frac{1}{2}{f}_1-\frac{1}{2}-\ln r_1)(-\frac{1}{2}{f}_1+\frac{1}{2}+\ln r_2)+f_2\ln \frac{r_2}{r_1}]}{4\mathrm{K\&pi;}(r_2^2-r_1^2)L\&CenterDot;\ln \frac{r_2}{r_1}}---\left(15\right)$

$-\frac{(\frac{1}{2}{f}_1-\frac{1}{2}-\ln r_1)(-\frac{1}{2}{f}_1+\frac{1}{2}+{\ln r}_2)}{2\mathrm{K\&pi;L}\&CenterDot;\ln \frac{r_2}{r_1}}$

Wherein, L is the axial length of circle wall cylinder body.

3. a kind of radially heat supply network network modeling method of heat conduction steady temperature maximum value of confirming according to claim 1 is characterized in that the decision content upper limit Δ described in the step 2 _MaxWith the lower limit Δ _MinSpecifically:

After the structure of justifying the wall cylinder body, material were confirmed, temperature maximum point was only relevant with the difference of the inside and outside surface temperature of justifying the wall cylinder body, then has:

$r_2^2=\frac{2k}{\stackrel{\&CenterDot;}{g}}\&CenterDot;\frac{1}{\ln \frac{r_2}{r_1}}[{\mathrm{\&Delta;}}_{\max }+\frac{\stackrel{\&CenterDot;}{g}}{4k}(r_2^2-r_1^2)]---\left(16\right)$

$r_1^2=\frac{2k}{\stackrel{\&CenterDot;}{g}}\&CenterDot;\frac{1}{\ln \frac{r_2}{r_1}}[{\mathrm{\&Delta;}}_{\min }+\frac{\stackrel{\&CenterDot;}{g}}{4k}(r_2^2-r_1^2)]---\left(17\right)$

Obtain decision content upper limit Δ by formula (16) and formula (17) _MaxWith decision content lower limit Δ _MinFor:

$\left\{\begin{array}{c}{\mathrm{\&Delta;}}_{\max }=|\frac{\stackrel{\&CenterDot;}{g}}{2k}\&CenterDot;\ln \frac{r_2}{r_1}\&CenterDot;r_2^2-\frac{\stackrel{\&CenterDot;}{g}}{4k}(r_2^2-r_1^2)|\\ {\mathrm{\&Delta;}}_{\min }=|\frac{\stackrel{\&CenterDot;}{g}}{2k}\&CenterDot;\ln \frac{r_2}{r_1}\&CenterDot;r_1^2-\frac{\stackrel{\&CenterDot;}{g}}{4k}(r_2^2-r_1^2)|\end{array}\right.---\left(18\right)$

Wherein,

Be the heat generation rate of circle wall cylinder body unit volume, k is the heat-conduction coefficient of circle wall cylinder block material, r ₁, r ₂Be respectively the internal diameter and the external diameter of round wall cylinder body.

4. a kind of radially heat supply network network modeling method of heat conduction steady temperature maximum value of confirming according to claim 1 and 2 is characterized in that, the approximate compensation tache Δ T in the said step 3, specifically:

Temperature maximum T _MaxExpression formula be:

$T_{\max }=-\frac{1}{2}{c}_1+c_1\ln {(\frac{2k}{\stackrel{\&CenterDot;}{g}}\&CenterDot;c_1)}^{\frac{1}{2}}+c_2---\left(6\right)$

Wherein, c ₁And c ₂Be two constants of diameter in the general solution that heat conduction steady-state heat balance equation is asked for,

Be the heat generation rate of circle wall cylinder body unit volume, k is the heat-conduction coefficient of circle wall cylinder block material, to wherein Do linear transformation, introduce parameter x, order

Xlnx is existed

Carry out the first-order linear match in the scope: xlnx ≈ f ₁X+f ₂, f ₁And f ₂Be the coefficient that the first-order linear match obtains, the error e that is produced by the first-order linear match is:

$e=\frac{1}{2}{c}_1\ln \left(\frac{2k}{\stackrel{\&CenterDot;}{g}}{c}_1\right)-(\frac{1}{2}{f}_1{c}_1+\frac{1}{2}\frac{\stackrel{\&CenterDot;}{g}}{2k}{f}_2)---\left(19\right)$

Wherein: $c_1=\frac{1}{\mathrm{Ln}\frac{r_2}{r_1}}[(T_2-T_1)+\frac{\stackrel{\&CenterDot;}{g}}{4k}(r_2^2-r_1^2)].$

Compensation tache Δ T=e then.

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