2013年美国大学生数学建模大赛A题论文.pdf

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1、For office use onlyT1 _T2 _T3 _T4 _TeamControl NumberFor office use onlyF1 _F2 _2259922599Problem ChosenF3 _F4 _A A20132013MathematicalMathematical ContestContest inin ModelingModeling (MCM/ICM)(MCM/ICM) SummarySummary SheetSheet(Attach a copy of this page to your solution paper.)HeatHeat RadiationR

2、adiation inin TheThe OvenOvenHeat distribution of pans in the oven is quite different from each other, whichdepends on their shapes. Thus, our model aims at two goals. One is to analyze theheat distri- bution in different ovens based on the locations of electrical heatingcubes. Further- more, a seri

3、es of heat distribution which varies from circular pans torectangular pans could be got easily. The other is to optimize the pans placing, inorder to choose a best way to maximize the even heat and the number of pans at thesame time.Mathematically speaking, our solution consists of two models, analy

4、zing andoptimi- zing. In part one, our whole-local approach shows the heat distribution ofevery pan. Firstly, we use the Stefan-Boltzmann law and Fourier theorem todescribe the heat distribution in the air around the electrical heating tube. And then,based on plane in- tercept method and simplified

5、Monte Carlo method, the heatdistribution of different shapes of pans is obtained. Finally, we explain thephenomenon that the corners of a pan always get over heated with water wavesstirring by analogy. In part two, our discretize- convert approach optimizes theshape and number of the pans. Above all

6、, we discre- tize the side length of the oven,so that the number and the average heat of the pans vary linearly. In the end, theabstract weight P is converted into a specific length, in order to reach a compromisebetween the two factors.Specially, we create a unique method to convert the variables f

7、rom the whole spaceto the local section. The special method allows us to draw the heat distribution ofevery single section in the oven. The algorithm we create does a great job inflexibility,which can be applied to all shapes of pans.Typea summary of your results on this page. Do not includethe name

8、 of your school, advisor, or team members on this page.HeatHeat RadiationRadiation inin TheThe OvenOvenSummarySummaryHeat distribution of pans in the oven is quite different from each other, which dependson their shapes. Thus, our model aims at two goals. One is to analyze the heat distri-bution in

9、different ovens based on the locations of electrical heating cubes. Further-more, a series of heat distribution which varies from circular pans to rectangular panscould be got easily.The other is to optimize the pans placing, in order to choose a bestway to maximize the even heat and the number of p

10、ans at the same time.Mathematically speaking, our solution consists of two models, analyzing and optimi-zing. In part one, our whole-local approach shows the heat distribution of every pan.Firstly,we use the Stefan-Boltzmann law and Fourier theorem to describe the heatdistribution in the air around

11、the electrical heating tube. And then, based on plane in-tercept method and simplified Monte Carlo method, the heat distribution of differentshapes of pans is obtained. Finally,we explain the phenomenon that the corners of apan always get over heated with water waves stirring by analogy.In part two,

12、 ourdiscretize- convert approach optimizes the shape and number of the pans.Above all,we discre- tize the side length of the oven, so that the number and the average heat ofthe pans vary linearly.In the end, the abstract weight P is converted into a specificlength, in order to reach a compromise bet

13、ween the two factors.Specially,we create a unique method to convert the variables from the whole space tothe local section. The special method allows us to draw the heat distribution of everysingle section in the oven. The algorithm we create does a great job in flexibility,which can be applied to a

14、ll shapes of pans.KeywordsKeywords:Monte Carlodiscretizationthermal radiationsectionheat distributionTeam#22599Page 1 of 23IntroductionIntroductionMany studies on heat conduction wasted plenty of time in solving the partialdifferential equations, since its difficult to solve even for computers. We t

15、urn toanother way to work it out. Firstly, we study the heat radiation instead of heatconduction to keep away from the sophisticated partial differential equations. Then,we create a unique method to convert every variable from the whole space to section.In other words, we work everything out in heat

16、 radiation and convert them into heatcontradiction.AssumptionsAssumptionsWemake the following assumptions about the distribution of heat in this paper.Initially two racks in the oven, evenly spaced.When heating the electrical heating tubes, the temperature of which changes fromroom temperature to th

17、e desired temperature. It takes such a short time that we canignore it. Different pans are made in same material, so they have the same rate of heatconduction.The inner walls of the oven are blackbodies. The pan is a gray body.The inner wallsof the oven absorb heat only and reflect no heat.The heat

18、can only be reflected once when rebounded from the pan.HeatHeat DistributionDistribution ModelModelOur approach involves four steps:Use the Fourier theorem to calculate the loss energy when energy beams arespread in the medium. So we can get the heat distribution around each electricalheating tube.

19、The heat distribution of the entire space could be go where the heat oftwo electrical heating tubes cross together.When different shapes of the pans are inserted into the oven, the heat map of theentire space is crossed by the section of the pan. Thus, the heat map of every singlepan is obtained.Est

20、ablish a suitable model to get the reflectivity of every single point on the pan withthe simplified Monte Carlo method.And then, a final heat distribution map of thepan without reflection loss is obtained.Arealistic conclusion is drawn due to the results of our model compared with waterwave propagat

21、ion phenomena.Team#22599Page 2 of 23First of all, the paper will give a description of the initial energy of the electrical hea-ting tube. Wesee it as a blackbody who reflects no heat at all. Electromagnetic know-ledge shows that wavelength of the heat rays ranges from 101umto 102umasshown below1:Fi

22、gure 1.Figure 2.Weapply the Stefan-Boltzmanns law2 whose solution isc15Eb=c2/(T)e1c15Eb=Ebd=c2/(T)de100WhereEbmeans the ability of blackbody to radiate.c1andc2are constants.Obviously,the initial energy of a black body isEb0= 3.2398e+012(wm2) .Combine Figure 1 with Figure 2, we integrate (1) from1to2

23、to get the equationas follow:2(1)(2)Eb(12)=Ebd1(3)Team#22599Page 3 of 23Figure 3.From Figure 3, it can be seen how the power of radiation varies with wavelength.Secondly,based on the Fourier theorem, the relation between heat and the distancefrom the electrical heating tubes is:dt(4)Q= SdxWhereQis t

24、he power of heat (J/s=W),Sis the area where the energy beamdtradiates (m2),represents the temperature gradient along the direction of energydxbeam. 3It is known that the energy becomes weaker as the distance becomes larger.Accordingto the fact we know:dQ(5)=dxWhereis the rate of energy changing.We a

25、ssume that the desired temperature of electrical heating tube is 500k. With thetwo equations, the distribution of heat is shown as follow:Figure 4. (a)Figure 4. (b)Team#22599Page 4 of 23In order to draw the map of heat distribution in the oven, we use MATLABto workon the complicated algorithm. The r

26、elation between the power of heat and the distanceis shown in Figure 4(a). The relation between temperature and distance is presented inFigure 4(b). The spreading direction of energy beam is presented in Figure 5.Figure 5.The shape of electrical heating tube is irregular. The heat distribution of a

27、singleelectrical heating tube can be draw in 3D space with MATLAB.The picture is shownin Figure 6. After superimposing, the total heat distribution of two tubes is shownbelow in Figure 7 and Figure 8.Figure 6.Figure 7.Team#22599Page 5 of 23Figure 8.The pictures above show the energy in an oven with

28、no pan. We put in a rectangularpan whose area isA, and intercept the maps with MATLAB. The result is show inFigure 9.Figure 9.Figure 10.Put in a circular pan to intercept the maps, whose area isA, also. The distribution ofheat is shown in Figure 11.Team#22599Page 6 of 23Figure 11.When put in a pan i

29、n transition shape, which is neither rectangular nor circular. Thearea of it isA, also. The heat distribution on such a pan is shown as follow:Figure 12(a)Team#22599Page 7 of 23Figure 12(b).Figure 13.Next, learning from the Monte Carlo simulation4, a model is established to getobtain the reflectivit

30、y.Wegenerate a random number between 0 and 1 to determine ifthe energy beam on certain point is reflected.Firstly,to demonstrate the question better , we construct a simple model:Team#22599Page 8 of 23Figure 14.Whereis the viewing angle from electrical heating tube to the pan.R=the proportion of the

31、 beams radiated to the pan. What is more, we assume the total beam isM1. Ideally, the number of absorption is. Then, each element of the pan is seen as a grid point. Each grid point canM1360generate a-M1-random-number vector between 0 and 1 in MATLAB.360 After MATLAB simulating, the number of beams

32、decreased byM2, due to the360M2reflection. So we define a probability=to describe the number of beamsM1reflected.The conclusion is : IfR, the energy beam is absorbed. IfR, the energy beam is reflected.5Based on the analysis above, our model get a final result of heat-distribution on thepan as shown

33、below:is360Figure 15(a)Team#22599Page 9 of 23Figure 15(b).The conclusion is known that the closer the shape of pans is to circle, the more evenlythe heat is distributed. Moreover, the phenomenon that the corners always get overheated can be explained by water wave propagation in different containers

34、.When there is a fluctuation in the center of the water, the ripples will fluctuate andspread in concentric circles, as shown in Figure 16. The fluctuation stirs waves upwhen contacting the borderlines. Compared with the waves with one boundary, thewaves in corner make a higher amplitude.The thermal

35、 conduction on the pan is exactly the inverse process of the wavespropagation. The range of thermal motion is much smaller than it on the side. Thatswhy the corners is easy to get over heated.In order to make the heat evenly distributed on the pan, the sides of the pan should beas few as possible. T

36、herefore, if nothing is considered about the utilization of space, acircle pan is the best choice.Figure 16, the water waves propagation6According to the analysis above and Figure 7, the phenomenon shows that the heatconduction is similar to water waves propagation. So it is proved that heatTeam#225

37、99Page 10 of 23concentrates in the four corners of the rectangular pan.TheThe SuperSuper PanPan ModelModelAssumptions The width of the oven (W) is100mm, the length isL. There are three pans at most in vertical direction. Each pansarea isA.The first part.Calculate the maximum number of pans in the ov

38、en. Different shapes of pan havedifferent heat distribution which affects the number of pans, judging from theprevious solution. According to the conclusion in the first model, the heat isdistributed the most evenly on a circular pan rather than a rectangular one. However,the rectangular pans make f

39、uller use of the space the space than circular ones. Bothfactors considered, a polygonal pan is chosen.A circle can be regard as a polygon whose number of boundaries tends to infinity.Except for rectangle, only regular hexagon and equilateral triangle can be closelyplaced. Because of the edges of eq

40、uilateral triangle, heat dissipation is worse thanrectangle. So, hexagonal pans are adopted after all the discussion.Considering the gaps near boundaries, we place the hexagonal pans closely attachedeach other on the long sideL. There are two kinds of programs as shown below.Program 1.Program 2.Team

41、#22599Page 11 of 23Obviously, Program2 is better than Program1 when considering space utilization. Soscheme 1 is adopted.Then, design a size of each hexagonal pan to make the highest space utilization. Withthe aim of utilization, hexagonal pans has to be placed contact closely with each otheron both

42、 sides. It is necessary to assume a aspect ratio of the oven to work out thenumber of pans(N).Assume that the side length of a regular hexagon isa, the length-width ratio of theoven isand Lis the increment in discretization. Because the number of panscan not change continuously whenn(m,m+1),m=1,2,3,

43、 the equations would beas follows.W5=a0 W1LL=L+kL03a;L=2L+kLL+kL10(6)WWW=L+kLN0=8,L0=WLWn= 3a 2n1N=N+3+1n= 2k1,k=1,2,3012Result:N=N+3nn= 2k,k=1,2,3202WhereN1represents the number of pans whennis odd,N2represents thenumber of pans whennis even. The specific number of pans is depended on thewidth-leng

44、th ratio of oven.The second part.Maximum the heat distribution of the pans. We define the average heat(H) as theratio of total heat and total area of the pans. Aiming to get the most average heat, weset the width-length ratio of the oven. Space utilization is not considered here.A conclusion is easy

45、 to draw from Figure 8 that a square area in the oven from150mmto 350mmin length shares the most heat evenly. So the pans should beTeam#22599Page 12 of 23placed mainly in this area. From model 1 we know that the corners of the oven are aptto gather heat. Besides, four more pans are added in the corn

46、ers to absorb more heat.Because heat absorbing is the only aim, there is no need to consider space utilization.Circular pans can distribute heat more evenly than any other shape due to model 1. Socircular pans are used in Figure 17.Figure 8.Figure 17.Weset the heat of the pans in the most heated are

47、a (the middle row) asQ. Pans in thecorners receive more heat but uneven theoretically.And the square of the four pans inthe corners is so small compared with the total square that we set the heat of the fourasQtoo. When the length of oven (L) increases, the number of pans increases too.It makes the

48、square of the gaps between pans bigger, meanwhile. If each pan has asame radius (r) and square (A), the equation about average heat, length-width ratioand number of pans would be (7).Team#22599Page 13 of 231 3r=W4 5A=r20 W1LL=L+kLL= 2rL+kLL+kL0 1WWW=L+kLN0= 7;L0=Wn=LW2rN=N0+nk=1,2,3.(7)Here we get t

49、he most average heat (H):400QH=29WThe third part.We discussed two different plans in the previous parts of the paper. One is aimed toget the most average heat, while the other aimed to place the most pans. The twoplans are contradictory with each other, and can not be achieved together.Firstly, the

50、weight of plan 1 isPand the weight of plan 2 is 1P. Obviously, thiskind optimization has difficulty in solving and understanding. So we turn to anotherway to make it a easier and linear question. It has been set that the width of the ovenis a constantWand there should be three pans at most in vertic